Short Version: when we say that $(\pmb{q},\pmb{u}):TQ_{(q)}\to\mathbb{R}^{2n}$ are local coordinates for the tangent bundle of $Q$, which can be viewed as an embedded submanifold of a higher dimensional euclidean space, what is the extrinsic view of the coordinate basis vectors $\pmb{\partial}_{u^i}$ associated with the velocity-level coordinate functions, $u^i$?
Long Version: Let $Q\subset \mathbb{E}^{3N}$ be an $n$-dimensional configuration manifold for some classical system of $N$ particles in 3-space subject to some $K$ holonomic constraints such that we can describe the configuration (position of all the particles) by some $n=3N-K$ generalized coordinates, $\pmb{q}=(q^1,\dots, q^n)$, giving a (local) diffeomorphism $\pmb{q}:Q_{(q)}\to \mathbb{R}^{n}$. The following is how I understand the extrinsic view of the associated coordinate basis on $Q$, defined in terms of a basis for $\mathbb{E}^{3N}$.
Let $\vec{e}_{\alpha} \in\mathbb{E}^{3N}$ be any global non-accelerating basis (not necessarily orthonormal) with dual basis $\vec{\epsilon}^a \in(\mathbb{E}^{3N})^*$. We can write a ''position vector'' for a point on $Q$, viewed as embedded in $\mathbb{E}^{3N}$:
$$ \vec{r}(\pmb{q}) = r^a (\pmb{q}) \vec{e}_a \in Q \subset \mathbb{E}^{3N} \qquad, \qquad \pmb{q}=\pmb{q}(\vec{r}) \in \mathbb{R}^n \qquad,\qquad a=1,\dots ,3N $$
(I'm abusing notation; many would name the chart $\varphi:Q_{(q)}\to\mathbb{R}^n$ and write $\varphi(\vec{r})=\pmb{q}$ and $\vec{r}=\varphi^{-1}(\pmb{q})= r^a (\pmb{q}) \vec{e}_a$) where $r^a:\mathbb{E}^{3N}\to\mathbb{R},\;\vec{r}\mapsto\vec{\epsilon}^a\cdot\vec{r}$, are global linear coordinates for $\mathbb{E}^{3N}$. The $\pmb{q}$ coordinate basis is then simply
$$ \pmb{\partial}_{i} = \frac{\partial r^a}{\partial q^i} \vec{e}_a \in \mathfrak{X}(Q) \qquad,\qquad \mathbf{d}q^i = \frac{\partial q^i}{\partial r^a} \vec{\epsilon}^a \in \Omega^1(Q) \qquad,\qquad i=1,\dots , n \qquad (1) $$
Is the above correct so far? I have defined a vector field $\pmb{\partial}_{i}\in\mathfrak{X}(Q)$ as a linear combination of vectors on $\mathbb{E}^{3N}$ (with functions as coefficients), and similarly for the basis 1-forms. Is this appropriate for the extrinsic view of things? I feel like $\pmb{\partial}_{i}$ is not actually equal to a linear combination of $\vec{e}_a$ as I have above, but is instead isomorphic in some trivial way?
Now the induced coordinate basis on $TQ$: The above $\pmb{\partial}_i|_r = \frac{\partial r^a}{\partial q^i}\big|_r \vec{e}_a \in T_rQ$ is a basis for each tangent space, with dual basis $\mathbf{d}q^i|_r \in T^*_rQ$. A basis gives a linear coordinate chart for a vector space such that for any $\mathbf{v}_r \in T_rQ$ we can write $\mathbf{v}_r = (\mathbf{d}q^i \cdot \mathbf{v}_r)\pmb{\partial}_i=:v^i\pmb{\partial}_i$ (i should technically write $\mathbf{d}q^i|_r$ and $\pmb{\partial}_i|_r$). We can then define a local coordinate chart on tangent bundle by the ''tangent lift'' $T\pmb{q}=:(\pmb{q},\pmb{u}):TQ_{(q)}\to\mathbb{R}^{2n}$ as
$$ (\pmb{q},\pmb{u}):TQ_{(q)}\to\mathbb{R}^{2n} \quad,\quad (r,\mathbf{v}_r) \mapsto (\pmb{q}(r), \pmb{u}(r,\mathbf{v}_r) ) = ( q^1(r),\dots,q^n(r), v^1,\dots, v^n) \quad,\quad u^i(r,\mathbf{v}_r) = \mathbf{d}q^i|_r\cdot\mathbf{v}_r =:v^i $$
That is, $q^i\in\mathcal{F}(Q)$ is the same as before (although now treated as $q^i\circ \tau\in\mathcal{F}(TQ)$ where $\tau:TQ\to Q$ is the projection) and $u^i\in\mathcal{F}(TQ)$ just gives the components of any tangent vector in the $\pmb{q}$ coordinate basis from Eq(1).
My Main Question: Now, for some vector field $\mathbf{X}\in\mathfrak{X}(TQ)$, we can write it in the local basis for the above coordinates $(\pmb{q},\pmb{u})$ as
$$ \mathbf{X} = V^i \pmb{\partial}_{q^i} \,+\, W^i \pmb{\partial}_{u^i} \; \in\mathfrak{X}(TQ) $$ where $V^i,W^i$ are just names I've given the components. As I understand it, if I re-name the $q^i$ coordinate basis vectors in Eq (1) as $\mathbf{e}_{q^i}=\frac{\partial r^a}{\partial q^i} \vec{e}_a\in\mathfrak{X}(Q)$, then the above $\pmb{\partial}_{q^i}\in\mathfrak{X}(TQ)$ can essentially be viewed as $\pmb{\partial}_{q^i} = \mathbf{e}_{q^i} \oplus \mathbf{0}$ (am I correct about this?). My main question is, what is the extrinsic interpretation of the the above $\pmb{\partial}_{u^i}$ basis vectors? Is there a way to define/view them in terms of the $\vec{e}_a\in\mathbb{E}^{3N}$ as in Eq(1)? Or in terms of the $\mathbf{e}_{q^i}\in\mathfrak{X}(Q)$?
Anther question: Same as above but for the basis 1-forms $\mathbf{d}q^i\in\Omega^1(TQ)$ and $\mathbf{d}u^i\in\Omega^1(TQ)$. If I re-name the basis 1-forms from Eq(1) as $\pmb{\epsilon}^{q^i}\in\Omega^1(Q)$, then I think that I view the $\mathbf{d}q^i\in\Omega^1(TQ)$ (which is really $\mathbf{d}(q^i\circ \tau)$) as $\pmb{\epsilon}^{q^i}\oplus \pmb{0}$? But what about the $\mathbf{d}u^i$?