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Looking at the proof in the book, I'd say that only the independence of complex structure on $T_v V$ from the point in the fiber is lacking a proof. This is indeed a bit technical, but here is a sketch:

Take $w \in T_v V$ and consider its lifts $w' \in T_{u'}(U_{\mathbb{R}}), \ w'' \in T_{u''}(U_{\mathbb{R}})$ for two points $u', u''$ in the fiber $\phi^{-1}(v)$. We need to show that $T\phi (Iw')=T\phi (Iw'')$ so that we can define $Iw$ by choosing a lift $w'$ and declaring $Iw=T\phi (I w')$ independently of the choice of the lift.

The fiber $\phi^{-1}(v)$ is a smooth submanifold of $U$ which is tangent to the holomorphic distribution $E$, and a smooth submanifold of a complex manifold $U$ is complex submanifold if and only if its tangent space at each point is a complex subspace of $T_u(U)$. This is satisfied, so $\phi^{-1}(v)$ is a complex submanifold of $U$. Shrinking $U$ if necessary, we can assume that $U$ is biholomorphic to a polydisk where $\phi^{-1}(v)$ is given by $z_{k+1}=...=z_{n}=0$ and points have coordinates $u'=(0,...0)$ and $u''=(\overline{a_1}, ..., \overline{a_k},0,..0)$. Consider a constant holomorphic vector field on $\phi^{-1}(v)$ given by $$\overline{a_1} \frac{\partial}{\partial z_1}|_a+...+\overline{a_k}\frac{\partial}{\partial z_k}|_a.$$ By construction it is a section of our holomorphic distribution $E$ and therefore can be written as $$f_1(a)X_1|_{(a,0)}+...+f_k(a)X_k|_{(a,0)}$$for some local frame $X_1, ..., X_k$ for $E$ on $U$ and functions $f_i(a)$, holomorphic on the fiber $\phi^{-1}(v)$. Trivially extend it to holomorphic section of $E$ on the whole $U$ by $$X|_{(a, b)}=f_1(a)X_1|_{(a,b)}+...+f_k(a)X_k|_{(a,b)}.$$

Now consider the flow of $X$ as a real vector field (using the identification of $TU$ with $T(U_{\mathbb{R}}$)). It is a fact (for example, see here or here) that at each time $t=t_0$ the flow $\Phi_{t_0}$ is a biholomorphism (on its domain of definition). Note that the curve $$\theta: [0,1] \to U$$ given by $$t \mapsto (\overline{a_1}t, ..., \overline{a_k} t, 0 ,...,0)$$is integral for $X$ and connects point $u'$ with point $u''$, so the flow $\Phi_{t=1}$ is defined in the neighborhood of $u'$ and maps to a neighborhood of $u''$ biholomorphically.

Now, recall the points in the holomorphic tangent space $T_u'(U)$ are represented by germs of complex curves $[z \mapsto \gamma(z)]$, while points of the smooth tangent space $T_u'(U_{\mathbb{R}})$ are represented by germs of smooth curves $[t \mapsto \theta(t)]$ and the identification (as real vector spaces) $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ simply goes like this $$[z \mapsto \gamma(z)] \mapsto [t \mapsto \gamma(t)].$$ Multiplication by $i$ acts on $T_{u'}(U)$ like $$i \cdot [z \mapsto \gamma(z)]=[z \mapsto \gamma(iz)],$$while the almost complex structure on $T_{u'}(U_{\mathbb{R}})$ induced by $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ acts by $$I[t \mapsto \gamma (t)]=[t \mapsto \gamma(it)].$$

So choose a complex germ $[z \mapsto \gamma(z)]$, such that the corresponding real curve $[t \mapsto \gamma (t)]$ represents the lift $w'$. Applying $\phi_{t=1}$, we get a holomorphic germ $[z \mapsto \Phi_{t=1} (\gamma (z))]\in T_{u''}(U)$. We claim that the inderlying smooth curve $[t \mapsto \Phi_{t=1} (\gamma (t))]$ is again a lift of $w$ so can be taken to be $w''$. For that consider the map $$(s, t) \mapsto \Phi_s (\gamma (t)).$$ For fixed $t$ the curve is integral for the vector field $X$, tangent to the distribution $E$, so $$\phi \circ \Phi_s \circ \gamma (t)$$ is a constant function of $s$ for fixed $t$. So $$w=T \phi (w')=[\phi (\gamma (t)]=[\phi \circ \Phi_{t=0} \circ \gamma (t)]= [\phi \circ \Phi_{t=1} \circ \gamma (t)]=T \phi [\Phi_{t=1} (\gamma (t))],$$as claimed.

Finally, since the construction was essentially holomorphic, it respects the almost complex structure! The vectors $$I w' = [t \mapsto \gamma (it)], \ \ Iw''=[t \mapsto \Phi_{t=1} (\gamma (it))$$are mapped to the same vector by $T\phi$, because we can repeat the argument above to show that $$(s,t) \mapsto \phi \circ \Phi_s \circ \gamma(it)$$is independent of $s$.

If points $u'$ and $u''$ do not lie in a single submanifold chart, one can repeat this construction several times along some path connecting $u'$ and $u'$ inside the fiber.

PS. I agree that probably proving the holomorphic Frobenius theorem would be easier than taking this route through the real Frobenius theorem. But this proof is instructive of the beautiful interplay between holomorphic and smooth geometries, underlying a complex manifold.

Looking at the proof in the book, I'd say that only the independence of complex structure on $T_v V$ from the point in the fiber is lacking a proof. This is indeed a bit technical, but here is a sketch:

Take $w \in T_v V$ and consider its lifts $w' \in T_{u'}(U_{\mathbb{R}}), \ w'' \in T_{u''}(U_{\mathbb{R}})$ for two points $u', u''$ in the fiber $\phi^{-1}(v)$. We need to show that $T\phi (Iw')=T\phi (Iw'')$ so that we can define $Iw$ by choosing a lift $w'$ and declaring $Iw=T\phi (I w')$ independently of the choice of the lift.

The fiber $\phi^{-1}(v)$ is a smooth submanifold of $U$ which is tangent to the holomorphic distribution $E$, and a smooth submanifold of a complex manifold $U$ is complex submanifold if and only if its tangent space at each point is a complex subspace of $T_u(U)$. This is satisfied, so $\phi^{-1}(v)$ is a complex submanifold of $U$. Shrinking $U$ if necessary, we can assume that $U$ is biholomorphic to a polydisk where $\phi^{-1}(v)$ is given by $z_{k+1}=...=z_{n}=0$ and points have coordinates $u'=(0,...0)$ and $u''=(\overline{a_1}, ..., \overline{a_k},0,..0)$. Consider a constant holomorphic vector field on $\phi^{-1}(v)$ given by $$\overline{a_1} \frac{\partial}{\partial z_1}|_a+...+\overline{a_k}\frac{\partial}{\partial z_k}|_a.$$ By construction it is a section of our holomorphic distribution $E$ and therefore can be written as $$f_1(a)X_1|_{(a,0)}+...+f_k(a)X_k|_{(a,0)}$$for some local frame $X_1, ..., X_k$ for $E$ on $U$ and functions $f_i(a)$, holomorphic on the fiber $\phi^{-1}(v)$. Trivially extend it to holomorphic section of $E$ on the whole $U$ by $$X|_{(a, b)}=f_1(a)X_1|_{(a,b)}+...+f_k(a)X_k|_{(a,b)}.$$

Now consider the flow of $X$ as a real vector field (using the identification of $TU$ with $T(U_{\mathbb{R}}$)). It is a fact (for example, see here or here) that at each time $t=t_0$ the flow $\Phi_{t_0}$ is a biholomorphism (on its domain of definition). Note that the curve $$\theta: [0,1] \to U$$ given by $$t \mapsto (\overline{a_1}t, ..., \overline{a_k} t, 0 ,...,0)$$is integral for $X$ and connects point $u'$ with point $u''$, so the flow $\Phi_{t=1}$ is defined in the neighborhood of $u'$ and maps to a neighborhood of $u''$ biholomorphically.

Now, recall the points in the holomorphic tangent space $T_u'(U)$ are represented by germs of complex curves $[z \mapsto \gamma(z)]$, while points of the smooth tangent space $T_u'(U_{\mathbb{R}})$ are represented by germs of smooth curves $[t \mapsto \theta(t)]$ and the identification (as real vector spaces) $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ simply goes like this $$[z \mapsto \gamma(z)] \mapsto [t \mapsto \gamma(t)].$$ Multiplication by $i$ acts on $T_{u'}(U)$ like $$i \cdot [z \mapsto \gamma(z)]=[z \mapsto \gamma(iz)],$$while the almost complex structure on $T_{u'}(U_{\mathbb{R}})$ induced by $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ acts by $$I[t \mapsto \gamma (t)]=[t \mapsto \gamma(it)].$$

So choose a complex germ $[z \mapsto \gamma(z)]$, such that the corresponding real curve $[t \mapsto \gamma (t)]$ represents the lift $w'$. Applying $\phi_{t=1}$, we get a holomorphic germ $[z \mapsto \Phi_{t=1} (\gamma (z))]\in T_{u''}(U)$. We claim that the inderlying smooth curve $[t \mapsto \Phi_{t=1} (\gamma (t))]$ is again a lift of $w$ so can be taken to be $w''$. For that consider the map $$(s, t) \mapsto \Phi_s (\gamma (t)).$$ For fixed $t$ the curve is integral for the vector field $X$, tangent to the distribution $E$, so $$\phi \circ \Phi_s \circ \gamma (t)$$ is a constant function of $s$ for fixed $t$. So $$w=T \phi (w')=[\phi (\gamma (t)]=[\phi \circ \Phi_{t=0} \circ \gamma (t)]= [\phi \circ \Phi_{t=1} \circ \gamma (t)]=T \phi [\Phi_{t=1} (\gamma (t))],$$as claimed.

Finally, since the construction was essentially holomorphic, it respects the almost complex structure! The vectors $$I w' = [t \mapsto \gamma (it)], \ \ Iw''=[t \mapsto \Phi_{t=1} (\gamma (it))$$are mapped to the same vector by $T\phi$, because we can repeat the argument above to show that $$(s,t) \mapsto \phi \circ \Phi_s \circ \gamma(it)$$is independent of $s$.

If points $u'$ and $u''$ do not lie in a single submanifold chart, one can repeat this construction several times along some path connecting $u'$ and $u'$ inside the fiber.

PS. I agree that probably proving the holomorphic Frobenius theorem from the first principles of analysis of complex variables would be easier than taking this route through the real Frobenius theorem. But this proof is instructive of the beautiful interplay between holomorphic and smooth geometries, underlying a complex manifold.

Looking at the proof in the book, I'd say that only the independence of complex structure on $T_v V$ from the point in the fiber is lacking a proof. This is indeed a bit technical, but here is a sketch:

Take $w \in T_v V$ and consider its lifts $w' \in T_{u'}(U_{\mathbb{R}}), \ w'' \in T_{u''}(U_{\mathbb{R}})$ for two points $u', u''$ in the fiber $\phi^{-1}(v)$. The fiber $\phi^{-1}(v)$ is a smooth submanifold of $U$ which is tangent to the holomorphic distribution $E$, and a smooth submanifold of a complex manifold $U$ is complex submanifold if and only if its tangent space at each point is a complex subspace of $T_u(U)$. This is satisfied, so $\phi^{-1}(v)$ is a complex submanifold of $U$. Shrinking $U$ if necessary, we can assume that $U$ is biholomorphic to a polydisk where $\phi^{-1}(v)$ is given by $z_{k+1}=...=z_{n}=0$ and points have coordinates $u'=(0,...0)$ and $u''=(\overline{a_1}, ..., \overline{a_k},0,..0)$. Consider a constant holomorphic vector field on $\phi^{-1}(v)$ given by $$\overline{a_1} \frac{\partial}{\partial z_1}|_a+...+\overline{a_k}\frac{\partial}{\partial z_k}|_a.$$ By construction it is a section of our holomorphic distribution $E$ and therefore can be written as $$f_1(a)X_1|_{(a,0)}+...+f_k(a)X_k|_{(a,0)}$$for some local frame $X_1, ..., X_k$ for $E$ on $U$ and functions $f_i(a)$, holomorphic on the fiber $\phi^{-1}(v)$. Trivially extend it to holomorphic section of $E$ on the whole $U$ by $$X|_{(a, b)}=f_1(a)X_1|_{(a,b)}+...+f_k(a)X_k|_{(a,b)}.$$

Now consider the flow of $X$ as a real vector field (using the identification of $TU$ with $T(U_{\mathbb{R}}$)). It is a fact (for example, see here or here) that at each time $t=t_0$ the flow $\Phi_{t_0}$ is a biholomorphism (on its domain of definition). Note that the curve $$\theta: [0,1] \to U$$ given by $$t \mapsto (\overline{a_1}t, ..., \overline{a_k} t, 0 ,...,0)$$is integral for $X$ and connects point $u'$ with point $u''$, so the flow $\Phi_{t=1}$ is defined in the neighborhood of $u'$ and maps to a neighborhood of $u''$ biholomorphically.

Now, recall the points in the holomorphic tangent space $T_u'(U)$ are represented by germs of complex curves $[z \mapsto \gamma(z)]$, while points of the smooth tangent space $T_u'(U_{\mathbb{R}})$ are represented by germs of smooth curves $[t \mapsto \theta(t)]$ and the identification (as real vector spaces) $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ simply goes like this $$[z \mapsto \gamma(z)] \mapsto [t \mapsto \gamma(t)].$$ Multiplication by $i$ acts on $T_{u'}(U)$ like $$i \cdot [z \mapsto \gamma(z)]=[z \mapsto \gamma(iz)],$$while the almost complex structure on $T_{u'}(U_{\mathbb{R}})$ induced by $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ acts by $$I[t \mapsto \gamma (t)]=[t \mapsto \gamma(it)].$$

So choose a complex germ $[z \mapsto \gamma(z)]$, such that the corresponding real curve $[t \mapsto \gamma (t)]$ represents the lift $w'$. Applying $\phi_{t=1}$, we get a holomorphic germ $[z \mapsto \Phi_{t=1} (\gamma (z))]\in T_{u''}(U)$. We claim that the inderlying smooth curve $[t \mapsto \Phi_{t=1} (\gamma (t))]$ is again a lift of $w$ so can be taken to be $w''$. For that consider the map $$(s, t) \mapsto \Phi_s (\gamma (t)).$$ For fixed $t$ the curve is integral for the vector field $X$, tangent to the distribution $E$, so $$\phi \circ \Phi_s \circ \gamma (t)$$ is a constant function of $s$ for fixed $t$. So $$w=T \phi (w')=[\phi (\gamma (t)]=[\phi \circ \Phi_{t=0} \circ \gamma (t)]= [\phi \circ \Phi_{t=1} \circ \gamma (t)]=T \phi [\Phi_{t=1} (\gamma (t))],$$as claimed.

Finally, since the construction was essentially holomorphic, it respects the almost complex structure! The vectors $$I w' = [t \mapsto \gamma (it)], \ \ Iw''=[t \mapsto \Phi_{t=1} (\gamma (it))$$are mapped to the same vector by $T\phi$, because we can repeat the argument above to show that $$(s,t) \mapsto \phi \circ \Phi_s \circ \gamma(it)$$is independent of $s$.

If points $u'$ and $u''$ do not lie in a single submanifold chart, one can repeat this construction several times along some path connecting $u'$ and $u'$ inside the fiber.

PS. I agree that probably proving the holomorphic Frobenius theorem would be easier than taking this route through the real Frobenius theorem. But this proof is instructive of the beautiful interplay between holomorphic and smooth geometries, underlying a complex manifold.

Looking at the proof in the book, I'd say that only the independence of complex structure on $T_v V$ from the point in the fiber is lacking a proof. This is indeed a bit technical, but here is a sketch:

Take $w \in T_v V$ and consider its lifts $w' \in T_{u'}(U_{\mathbb{R}}), \ w'' \in T_{u''}(U_{\mathbb{R}})$ for two points $u', u''$ in the fiber $\phi^{-1}(v)$. We need to show that $T\phi (Iw')=T\phi (Iw'')$ so that we can define $Iw$ by choosing a lift $w'$ and declaring $Iw=T\phi (I w')$ independently of the choice of the lift.

The fiber $\phi^{-1}(v)$ is a smooth submanifold of $U$ which is tangent to the holomorphic distribution $E$, and a smooth submanifold of a complex manifold $U$ is complex submanifold if and only if its tangent space at each point is a complex subspace of $T_u(U)$. This is satisfied, so $\phi^{-1}(v)$ is a complex submanifold of $U$. Shrinking $U$ if necessary, we can assume that $U$ is biholomorphic to a polydisk where $\phi^{-1}(v)$ is given by $z_{k+1}=...=z_{n}=0$ and points have coordinates $u'=(0,...0)$ and $u''=(\overline{a_1}, ..., \overline{a_k},0,..0)$. Consider a constant holomorphic vector field on $\phi^{-1}(v)$ given by $$\overline{a_1} \frac{\partial}{\partial z_1}|_a+...+\overline{a_k}\frac{\partial}{\partial z_k}|_a.$$ By construction it is a section of our holomorphic distribution $E$ and therefore can be written as $$f_1(a)X_1|_{(a,0)}+...+f_k(a)X_k|_{(a,0)}$$for some local frame $X_1, ..., X_k$ for $E$ on $U$ and functions $f_i(a)$, holomorphic on the fiber $\phi^{-1}(v)$. Trivially extend it to holomorphic section of $E$ on the whole $U$ by $$X|_{(a, b)}=f_1(a)X_1|_{(a,b)}+...+f_k(a)X_k|_{(a,b)}.$$

Now consider the flow of $X$ as a real vector field (using the identification of $TU$ with $T(U_{\mathbb{R}}$)). It is a fact (for example, see here or here) that at each time $t=t_0$ the flow $\Phi_{t_0}$ is a biholomorphism (on its domain of definition). Note that the curve $$\theta: [0,1] \to U$$ given by $$t \mapsto (\overline{a_1}t, ..., \overline{a_k} t, 0 ,...,0)$$is integral for $X$ and connects point $u'$ with point $u''$, so the flow $\Phi_{t=1}$ is defined in the neighborhood of $u'$ and maps to a neighborhood of $u''$ biholomorphically.

Now, recall the points in the holomorphic tangent space $T_u'(U)$ are represented by germs of complex curves $[z \mapsto \gamma(z)]$, while points of the smooth tangent space $T_u'(U_{\mathbb{R}})$ are represented by germs of smooth curves $[t \mapsto \theta(t)]$ and the identification (as real vector spaces) $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ simply goes like this $$[z \mapsto \gamma(z)] \mapsto [t \mapsto \gamma(t)].$$ Multiplication by $i$ acts on $T_{u'}(U)$ like $$i \cdot [z \mapsto \gamma(z)]=[z \mapsto \gamma(iz)],$$while the almost complex structure on $T_{u'}(U_{\mathbb{R}})$ induced by $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ acts by $$I[t \mapsto \gamma (t)]=[t \mapsto \gamma(it)].$$

So choose a complex germ $[z \mapsto \gamma(z)]$, such that the corresponding real curve $[t \mapsto \gamma (t)]$ represents the lift $w'$. Applying $\phi_{t=1}$, we get a holomorphic germ $[z \mapsto \Phi_{t=1} (\gamma (z))]\in T_{u''}(U)$. We claim that the inderlying smooth curve $[t \mapsto \Phi_{t=1} (\gamma (t))]$ is again a lift of $w$ so can be taken to be $w''$. For that consider the map $$(s, t) \mapsto \Phi_s (\gamma (t)).$$ For fixed $t$ the curve is integral for the vector field $X$, tangent to the distribution $E$, so $$\phi \circ \Phi_s \circ \gamma (t)$$ is a constant function of $s$ for fixed $t$. So $$w=T \phi (w')=[\phi (\gamma (t)]=[\phi \circ \Phi_{t=0} \circ \gamma (t)]= [\phi \circ \Phi_{t=1} \circ \gamma (t)]=T \phi [\Phi_{t=1} (\gamma (t))],$$as claimed.

Finally, since the construction was essentially holomorphic, it respects the almost complex structure! The vectors $$I w' = [t \mapsto \gamma (it)], \ \ Iw''=[t \mapsto \Phi_{t=1} (\gamma (it))$$are mapped to the same vector by $T\phi$, because we can repeat the argument above to show that $$(s,t) \mapsto \phi \circ \Phi_s \circ \gamma(it)$$is independent of $s$.

If points $u'$ and $u''$ do not lie in a single submanifold chart, one can repeat this construction several times along some path connecting $u'$ and $u'$ inside the fiber.

PS. I agree that probably proving the holomorphic Frobenius theorem would be easier than taking this route through the real Frobenius theorem. But this proof is instructive of the beautiful interplay between holomorphic and smooth geometries, underlying a complex manifold.

Looking at the proof in the book, I'd say that only the independence of complex structure on $T_v V$ from the point in the fiber is lacking a proof. This is indeed a bit technical, but here is a sketch:

Take $w \in T_v V$ and consider its lifts $w' \in T_{u'}(U_{\mathbb{R}}), \ w'' \in T_{u''}(U_{\mathbb{R}})$ for two points $u', u''$ in the fiber $\phi^{-1}(v)$. The fiber $\phi^{-1}(v)$ is a smooth submanifold of $U$ which is tangent to the holomorphic distribution $E$, and a smooth submanifold of a complex manifold $U$ is complex submanifold if and only if its tangent space at each point is a complex subspace of $T_u(U)$. This is satisfied, so $\phi^{-1}(v)$ is a complex submanifold of $U$. Shrinking $U$ if necessary, we can assume that $U$ is biholomorphic to a polydisk where $\phi^{-1}(v)$ is given by $z_{k+1}=...=z_{n}=0$ and points have coordinates $u'=(0,...0)$ and $u''=(\overline{a_1}, ..., \overline{a_k},0,..0)$. Consider a constant holomorphic vector field on $\phi^{-1}(v)$ given by $$\overline{a_1} \frac{\partial}{\partial z_1}|_a+...+\overline{a_k}\frac{\partial}{\partial z_k}|_a.$$ By construction it is a section of our holomorphic distribution $E$ and therefore can be written as $$f_1(a)X_1|_{(a,0)}+...+f_k(a)X_k|_{(a,0)}$$for some local frame $X_1, ..., X_k$ for $E$ on $U$ and functions $f_i(a)$, holomorphic on the fiber $\phi^{-1}(v)$. Trivially extend it to holomorphic section of $E$ on the whole $U$ by $$X|_{(a, b)}=f_1(a)X_1|_{(a,b)}+...+f_k(a)X_k|_{(a,b)}.$$

Now consider the flow of $X$ as a real vector field (using the identification of $TU$ with $T(U_{\mathbb{R}}$)). It is a fact (for example, see here) that at each time $t=t_0$ the flow $\Phi_{t_0}$ is a biholomorphism (on its domain of definition). Note that the curve $$\theta: [0,1] \to U$$ given by $$t \mapsto (\overline{a_1}t, ..., \overline{a_k} t, 0 ,...,0)$$is integral for $X$ and connects point $u'$ with point $u''$, so the flow $\Phi_{t=1}$ is defined in the neighborhood of $u'$ and maps to a neighborhood of $u''$ biholomorphically.

Now, recall the points in the holomorphic tangent space $T_u'(U)$ are represented by germs of complex curves $[z \mapsto \gamma(z)]$, while points of the smooth tangent space $T_u'(U_{\mathbb{R}})$ are represented by germs of smooth curves $[t \mapsto \theta(t)]$ and the identification (as real vector spaces) $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ simply goes like this $$[z \to \gamma(z)] \mapsto [t \to \gamma(t)].$$ Multiplication by $i$ acts on $T_{u'}(U)$ like $$i \cdot [z \mapsto \gamma(z)]=[z \mapsto \gamma(iz)],$$while the almost complex structure on $T_{u'}(U_{\mathbb{R}})$ induced by $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ acts by $$I[t \mapsto \gamma (t)]=[t \mapsto \gamma(it)].$$

So choose a complex germ $[z \mapsto \gamma(z)]$, such that the corresponding real curve $[t \mapsto \gamma (t)]$ represents the lift $w'$. Applying $\phi_{t=1}$, we get a holomorphic germ $[z \mapsto \Phi_{t=1} (\gamma (z))]\in T_{u''}(U)$. We claim that the inderlying smooth curve $[t \mapsto \Phi_{t=1} (\gamma (t))]$ is again a lift of $w$ so can be taken to be $w''$. For that consider the map $$(s, t) \mapsto \Phi_s (\gamma (t)).$$ For fixed $t$ the curve is integral for the vector field $X$, tangent to the distribution $E$, so $$\phi \circ \Phi_s \circ \gamma (t)$$ is a constant function of $s$ for fixed $t$. So $$w=T \phi (w')=[\phi (\gamma (t)]=[\phi \circ \Phi_{t=0} \circ \gamma (t)]= [\phi \circ \Phi_{t=1} \circ \gamma (t)]=T \phi [\Phi_{t=1} (\gamma (t))],$$as claimed.

Finally, since the construction was essentially holomorphic, it respects the almost complex structure! The vectors $$I w' = [t \mapsto \gamma (it)], \ \ Iw''=[t \mapsto \Phi_{t=1} (\gamma (it))$$are mapped to the same vector by $T\phi$, because we can repeat the argument above to show that $$(s,t) \mapsto \phi \circ \Phi_s \circ \gamma(it)$$is independent of $s$.

If points $u'$ and $u''$ do not lie in a single submanifold chart, one can repeat this construction several times along some path connecting $u'$ and $u'$ inside the fiber.

PS. I agree that probably proving the holomorphic Frobenius theorem would be easier than taking this route through the real Frobenius theorem. But this proof is instructive of the beautiful interplay between holomorphic and smooth geometries, underlying a complex manifold.

Looking at the proof in the book, I'd say that only the independence of complex structure on $T_v V$ from the point in the fiber is lacking a proof. This is indeed a bit technical, but here is a sketch:

Take $w \in T_v V$ and consider its lifts $w' \in T_{u'}(U_{\mathbb{R}}), \ w'' \in T_{u''}(U_{\mathbb{R}})$ for two points $u', u''$ in the fiber $\phi^{-1}(v)$. The fiber $\phi^{-1}(v)$ is a smooth submanifold of $U$ which is tangent to the holomorphic distribution $E$, and a smooth submanifold of a complex manifold $U$ is complex submanifold if and only if its tangent space at each point is a complex subspace of $T_u(U)$. This is satisfied, so $\phi^{-1}(v)$ is a complex submanifold of $U$. Shrinking $U$ if necessary, we can assume that $U$ is biholomorphic to a polydisk where $\phi^{-1}(v)$ is given by $z_{k+1}=...=z_{n}=0$ and points have coordinates $u'=(0,...0)$ and $u''=(\overline{a_1}, ..., \overline{a_k},0,..0)$. Consider a constant holomorphic vector field on $\phi^{-1}(v)$ given by $$\overline{a_1} \frac{\partial}{\partial z_1}|_a+...+\overline{a_k}\frac{\partial}{\partial z_k}|_a.$$ By construction it is a section of our holomorphic distribution $E$ and therefore can be written as $$f_1(a)X_1|_{(a,0)}+...+f_k(a)X_k|_{(a,0)}$$for some local frame $X_1, ..., X_k$ for $E$ on $U$ and functions $f_i(a)$, holomorphic on the fiber $\phi^{-1}(v)$. Trivially extend it to holomorphic section of $E$ on the whole $U$ by $$X|_{(a, b)}=f_1(a)X_1|_{(a,b)}+...+f_k(a)X_k|_{(a,b)}.$$

Now consider the flow of $X$ as a real vector field (using the identification of $TU$ with $T(U_{\mathbb{R}}$)). It is a fact (for example, see here or here) that at each time $t=t_0$ the flow $\Phi_{t_0}$ is a biholomorphism (on its domain of definition). Note that the curve $$\theta: [0,1] \to U$$ given by $$t \mapsto (\overline{a_1}t, ..., \overline{a_k} t, 0 ,...,0)$$is integral for $X$ and connects point $u'$ with point $u''$, so the flow $\Phi_{t=1}$ is defined in the neighborhood of $u'$ and maps to a neighborhood of $u''$ biholomorphically.

Now, recall the points in the holomorphic tangent space $T_u'(U)$ are represented by germs of complex curves $[z \mapsto \gamma(z)]$, while points of the smooth tangent space $T_u'(U_{\mathbb{R}})$ are represented by germs of smooth curves $[t \mapsto \theta(t)]$ and the identification (as real vector spaces) $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ simply goes like this $$[z \mapsto \gamma(z)] \mapsto [t \mapsto \gamma(t)].$$ Multiplication by $i$ acts on $T_{u'}(U)$ like $$i \cdot [z \mapsto \gamma(z)]=[z \mapsto \gamma(iz)],$$while the almost complex structure on $T_{u'}(U_{\mathbb{R}})$ induced by $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ acts by $$I[t \mapsto \gamma (t)]=[t \mapsto \gamma(it)].$$

So choose a complex germ $[z \mapsto \gamma(z)]$, such that the corresponding real curve $[t \mapsto \gamma (t)]$ represents the lift $w'$. Applying $\phi_{t=1}$, we get a holomorphic germ $[z \mapsto \Phi_{t=1} (\gamma (z))]\in T_{u''}(U)$. We claim that the inderlying smooth curve $[t \mapsto \Phi_{t=1} (\gamma (t))]$ is again a lift of $w$ so can be taken to be $w''$. For that consider the map $$(s, t) \mapsto \Phi_s (\gamma (t)).$$ For fixed $t$ the curve is integral for the vector field $X$, tangent to the distribution $E$, so $$\phi \circ \Phi_s \circ \gamma (t)$$ is a constant function of $s$ for fixed $t$. So $$w=T \phi (w')=[\phi (\gamma (t)]=[\phi \circ \Phi_{t=0} \circ \gamma (t)]= [\phi \circ \Phi_{t=1} \circ \gamma (t)]=T \phi [\Phi_{t=1} (\gamma (t))],$$as claimed.

Finally, since the construction was essentially holomorphic, it respects the almost complex structure! The vectors $$I w' = [t \mapsto \gamma (it)], \ \ Iw''=[t \mapsto \Phi_{t=1} (\gamma (it))$$are mapped to the same vector by $T\phi$, because we can repeat the argument above to show that $$(s,t) \mapsto \phi \circ \Phi_s \circ \gamma(it)$$is independent of $s$.

If points $u'$ and $u''$ do not lie in a single submanifold chart, one can repeat this construction several times along some path connecting $u'$ and $u'$ inside the fiber.

PS. I agree that probably proving the holomorphic Frobenius theorem would be easier than taking this route through the real Frobenius theorem. But this proof is instructive of the beautiful interplay between holomorphic and smooth geometries, underlying a complex manifold.