Find a specific countable atlas for a smooth submanifold with boundary Let $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary.$^1$
We know that there is a countable family $((\Omega_i,\phi_i))_{i\in I}$ of $k$-dimensional $C^1$-charts$^2$ of $M$ with $$M\subseteq\bigcup_{i\in I}\Omega_i.$$
We know that there is a $k$-dimensional boundary $C^1$-atlas$^1$ $((\Omega_i,\phi_i))_{i\in I}$ of $M$ for some $I\subseteq\mathbb N$.
Let $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$. Note that $(\mathbb H^k)^\circ=\mathbb R^{k-1}\times(0,\infty)$ and $\partial\mathbb H^k=\mathbb R^{k-1}\times\{0\}$.
Let $B$ denote the closed unit ball in $\mathbb R^k$, $B_+:=B\cap(\mathbb H^k)^\circ$ and $B_0:=B\cap\partial\mathbb H^k$.

Why can we choose $((\Omega_i,\phi_i))_{i\in I}$ such that the manifold interior$^3$ $\Omega_i^\circ$ is equal to $\phi_i^{-1}(B_+)$ and the manifold boundary $\partial\Omega_i$ is equal to $\phi_i^{-1}(B_0)$?


$^1$ i.e. each point of $M$ is locally $C^1$-diffeomorphic to $\mathbb H^k$.
If $E_i$ is a $\mathbb R$-Banach space and $B_i\subseteq E_i$, then $f:B_1\to E_2$ is called $C^1$-differentiable if $f=\left.\tilde f\right|_{B_1}$ for some $E_1$-open neighborhood $\Omega_1$ of $B_1$ and some $\tilde f\in C^1(\Omega_1,E_2)$ and $g:B_1\to B_2$ is called $C^1$-diffeomorphism if $g$ is a homeomorphism from $B_1$ onto $B_2$ and $g$ and $g^{-1}$ are $C^1$-differentiable.
$^2$ A $k$-dimensional $C^1$-chart of $M$ is a $C^1$-diffeomorphism from an open subset of $M$ onto an open subset of $\mathbb H^k$.
$^3$ $x\in M^\circ$ if and only if there is a $k$-dimensional $C^1$-chart $(\Omega,\phi)$ of $M$ such that $x\in\Omega$ and $\phi(\Omega)$ is $\mathbb R^k$-open.
$x\in\partial M$ if and only if there is a $k$-dimensional $C^1$-chart $(\Omega,\phi)$ of $M$ such that $x\in\Omega$ and $\phi(x)\in\partial\mathbb H^k$.
 A: 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 $\mathbb R^k$.
