Let $\phi :H\rightarrow M$ be a embedded submanifold of $M$. If I want to construct a smooth projection map $tan:\mathfrak X(M) \rightarrow \mathfrak X(H) $, one way is to use a Riemannian metric (whose existence is well known) to construct one (see [1] below). I want to understand this in a different way if possible.
Pointwise, I see the following happening:
At any point $p \in H$, consider $T_{\phi(p)}M$. Now, $\phi'(T_pH)$ is a linear subspace of $T_{\phi(p)}M$. By injectivity of $\phi'$, it is clear that $T_pH\cong \phi'(T_pH) $. Further, by standard linear algebra arguments and the definition of normal bundle, and above, one can easily say that: $$T_{\phi(p)}M \cong T_pH\oplus N_pH$$
Use the standard projectors $pr_i$ of the direct sum to construct the projector of the vector field as $$(tan(X))(p)=(\phi')^{-1}\circ pr_1(X(\phi(p)))$$ Ofcourse, the key caveat here is that we want the map to be smooth [2], and the problematic part is the $pr_1$ above. Now, I do not expect this map to be smooth, but I can't think of a counterexample to show that the above map is not smooth either.
[1] I know that given a Riemannian structure $g$ on $M$, one can induce one on $H$ as $h:=\phi^*g$. This yields a very natural smooth projection map as $tan:=h^\sharp \circ \phi^* \circ g^\flat$. But I am not interested in this.
[2] By smoothness of tan, I mean for all $X \in \mathfrak X(M)$, $tan(X) \in \mathfrak X(H)$ is smooth as a vector field of $H$.
I think you map $\tan$ is smooth. This can be checked locally, so let $p$ be an arbitrary point of $H$ and $U$ a sufficiently small open neighborhood of $p$. Take $V_1, \dots, V_n$ to be a local orthonormal frame of $T(U \cap H)$ (you can always obtain such a frame by starting with an arbitrary frame of $T(U \cap H)$ and applying Gram-Schmidt orthogonalization to it - locally it works for vector fields, too). First, the vector fields of this frame can be extended to orthogonal vector fields on $U$ (since $H \cap U$ is a embedded submanifold of $U$), and these can be extended to a orthogonal frame $V_1, \dots, V_m$ on $U$.
Now every vector field $X$ on $U$ can be expressed as a linear combination of $V_1, \dots V_m$ with smooth coefficients: $$ X = \sum_{k=1}^m f_k V_k $$ and it's easy to see that $$ \tan(X) = \sum_{k=1}^n f_k V_k $$ Since the $f_k$ are smooth, $\tan(X)$ is smooth.
Note that this also works for immersed manifolds.