Find a specific countable atlas for a smooth submanifold with boundary

When working with a (smooth or $C^k$) manifold $M$, implicitly you are saying that a (smooth or $C^k$) structure exists, that is, there exists some associated atlas ${ phi_i : V_i to mathbb R^k }_{i in I}$ corresponding to the (smooth or $C^k$) structure of interest. The key word is exists; there really isn't a canonical choice of atlas. Enter the notion of compatible atlases. We say another atlas ${psi_j : U_j to mathbb R^k}_{j in J}$ is compatible if $$psi_j circ phi_i^{-1} : phi_i (V_i) to mathbb R^k$$ is smooth for every $i in I$ and $j in J$. Philosophically speaking, this means both atlases are essentially the same. Mathematically, compatibility forms an equivalence relation on the collection of all atlases of a manifold, which each equivalence class being what is called a (smooth or $C^k$) structure.

The phrase "we can choose" translates to "we can pick a compatible atlas". Again, philosophically speaking, there is no difference in working with either of two compatible atlases. However, mathematically one atlas may be more convenient to work with than another. For instance, we might want the images $psi_j (U_j)$ to be balls to (what I like to call) "abuse" the local convex structure of Euclidean space. You see this a lot if you're working with Riemann surfaces.

So how does one arrive at such an atlas? Well, we know from Euclidean topology that open sets, for example $phi_i (V_i)^circ subseteq mathbb R^k$, can be written as countable unions of open balls $B_{i, j}$. This lets us define bijections $psi_{i, j} : phi_i^{-1} (B_{i, j}) to B_{i, j}$ by $psi_{i, j} := phi_i$.

So what about the boundary points? Well, we know that $phi_i (V_i)$ is open in $mathbb H^k$, so there exists a set $O subseteq mathbb R^k$ open in the Euclidean topology such that $$phi_i (V_i) = O cap mathbb H^k.$$ In particular, suppose $x in partial M cap V_i$, then $phi_i (x) in partial mathbb H^k$ and is an interior point of $O$. So we draw a ball around $phi_i (x)$ which is contained in $O$. Then we can draw a ball which is contained in this ball and $mathbb H^k$ with boundary point $phi_i (x)$, call it $B_{i, x}$. Then as with before we have a well-defined bijection $psi_{i, x} : phi_i^{-1} (B_{i, x}) to B_{i, x}$ given by $psi_{i, x} := phi_i$.

It's useful to draw a picture:

And now we're done! Since the $psi_{i, j}$ and $psi_{i, x}$ maps were defined using $phi_i$, collecting all of them together gives a (smooth or $C^k$) atlas. Of course this won't be countable since there are uncountably many boundary points $x in partial M$, however you can adapt the argument, instead of using balls tangent to $partial mathbb H$ you can use parallelepipeds (with rounded corners) and take a diffeomorphism sending it to a ball in $mathbb R^k$.