I have no idea what the proof in the book does. But, here is an argument.
Note that $H$ is an orthogonal projection ($H^2=H'=H$); as you say, it projects onto the column space of $X$. Similarly, $H_1$ is the orthogonal projection onto the column space of $X_1$, and $I-H_1$ is the projection onto the columns space of $X_2$. Let us call
$$
K=H_1+(I-H_1)X_2(X_2'(I-H_1)X_2)^{-1}X_2'(I-H_1).
$$
Since $H_1(I-H_1)=0$, it is very easy to check that $K^2=K$, and that $K'=K$. So $K$ is an orthogonal projection. Its range contains the column space of $X_1$, because if $x$ is in the column space of $X_1$ then $H_1x=x$, and so $Kx=x$. Note that
$$
(I-H_1)X_2(X_2'(I-H_1)X_2)^{-1}X_2'(I-H_1)
$$
is the orthogonal projection onto the column space of $(I-H_1)X_2$. So $K$ is the orthogonal sum of two orthogonal projections; to conclude that $K=H$ we need to show that the span of the column spaces of $X_1$ and $(I-H_1)X_2$ agrees with the column space of $X$.
If $y$ is in the column space of $X$, and $c_1,\ldots,c_m$ are the columns of $X$, we can write
$$
y
=(\alpha_1c_1+\cdots+\alpha_rc_r)+(\alpha_{r+1}d_{1}+\cdots+\alpha_sd_s),
$$
where $c_1,\ldots,c_r$ are the columns of $X_1$ and $d_{1},\ldots,d_s$ is an orthonormal basis obtained by applying Gram-Schmidt to those of $c_{r+1},\ldots,c_m$ that are linearly independent with $c_1,\ldots,c_r$. Then
$$
H_1y=\alpha_1c_1+\cdots+\alpha_rc_r
$$
and
$$
(I-H_1)y=\alpha_{r+1}d_{1}+\cdots+\alpha_sd_s.
$$
That is, $H_1y$ belongs to the span of the columns of $X_1$ and $(I-H_1)y$ belongs to the span of the columns of $(I-H_1)X_2$. Then
\begin{align}
Ky&=K(H_1y+(I-H_1)y)\\ \ \\
&=H_1^2y+(I-H_1)X_2(X_2'(I-H_1)X_2)^{-1}X_2'(I-H_1)\,(I-H_1)y\\ \ \\
&=H_1y+(I-H_1)y=y.
\end{align}
This shows that $K$ is the identity on the column space of $X$.
Conversely, if $y$ is orthogonal to the column space of $X$, then in particular $y$ is orthogonal to the column spaces of $X_1$ and $X_2$. Then $H_1y=0$. But then
$$
(I-H_1)X_2(X_2'(I-H_1)X_2)^{-1}X_2'(I-H_1)y=(I-H_1)X_2(X_2'(I-H_1)X_2)^{-1}X_2'y=0,
$$
the second equality since $X_2'y=0$ (because $y$ is orthogonal to the column space of $X_2$). Then we have that $Ky=0$.
We have shown that $K$ is the identity on the range of $H$ and zero on the orthogonal complement of the range of $H$. That is, for any $y$,
$$
Ky=K(Hy+(I-H)y)=Hy.
$$
so $K=H$.
