How do we obtain a chart for the sphere $\{x:\|x\|=1\}$ from a chart of the ball $\{x:\|x|\le1\}$?

Let $d\in\mathbb N$, $k\in\{1,\ldots,d\}$, $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary, $(\Omega,\phi)$ be a $k$-dimensional $C^1$-chart of $M$ (i.e. $\Omega$ is an open subset of $M$ and $\phi$ is a $C^1$-diffeomorphism from $\Omega$ onto an open subset of $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$) and $U:=\phi(\Omega)$.

We know that the manifold boundary $\partial\Omega$ is equal to $\partial M\cap\Omega$, $$\phi(\partial\Omega)=U\cap\partial\mathbb H^k,$$ where $\partial\mathbb H^k=\mathbb R^{k-1}\times\{0\}$, and $\left.\pi\circ\phi\right|_{\partial\Omega}$ is a $C^1$-diffeomorphism from $\partial\Omega$ onto the open subset $$(\pi\circ\phi)(\partial\Omega)=\{(u_1,\ldots,u_{k-1}):u\in U\text{ with }u_k=0\}\tag1$$ of $\mathbb R^{k-1}$, where $\pi$ denotes the canonical projection of $\mathbb R^k$ onto $\mathbb R^{k-1}$ with $\pi(\partial\mathbb H^{k-1})=\mathbb R^{k-1}$. So, $(\partial\Omega,\pi\circ\left.\phi\right|_{\partial\Omega})$ is a $(k-1)$-dimensional chart of $\partial M$.

Now consider the almost trivial example $k=d=3$ and $$M:=\{x\in\mathbb R^3:\|x\|\le1\}.$$ What would be a possible choice for $(\Omega,\phi)$?

I know how we can directly construct a $2$-dimensional chart of the sphere $\partial M=\{x\in\mathbb R^3:\|x\|=1\}$ (for example by steoreographic projections or graphs of the function $\mathbb R^2\ni x\mapsto\sqrt{1-\|x\|^2}$), but I would really like to know how the $2$-dimensional chart $(\partial\Omega,\pi\circ\left.\phi\right|_{\partial\Omega})$ of $\partial M$ looks like for a $3$-dimensional chart $(\Omega,\phi)$ of $M$.

By the way, what would happen if we replace $M$ by the open ball $\{x\in\mathbb R^3:\|x\|<1\}$? Since $M$ is open in this example, isn't $(\Omega,\phi)=(M,\operatorname{id}_M)$ a possible choice? I guess not and I'm missing something, since I really don't see why $(\partial\Omega,\pi\circ\left.\phi\right|_{\partial\Omega})$ would be a $2$-dimensional chart of $\partial M$ then ...

For $M = \{x \in \mathbb{R}^3 : \|x\| \leq 1\}$, one possible chart on the boundary is given by

$$ \Omega = \{(x_1,x_2,x_3) \in \mathbb{R}^3 : \|x\| \leq 1 \land x_3 > 0\} $$ $$ U = \{x \in \mathbb{R}^2 \times [0,\infty) : \|x\| < 1 \}$$ $$ \phi((x_1,x_2,x_3)) = \left(x_1, x_2, \sqrt{1-x_1^2-x_2^2} - x_3\right) $$

Since everywhere in $\Omega$ we have $x_1^2+x_2^2 < 1$, it's easy to see that $\phi$ is a bijection from $\Omega$ to $U$, maps $\partial \Omega$ into $\mathbb{R}^2 \times \{0\}$, and is $C^\infty$.

Then $\pi \circ \phi : \partial \Omega \to \mathbb{R}^2$ is found just by dropping the last coordinate:

$$ (\pi \circ \phi)((x_1, x_2, x_3)) = (x_1, x_2) $$

In fact for this particular chart, $\pi \circ \phi = \pi|_{\partial \Omega}$.

What would happen if we replace $M$ by the open ball $\{ x \in \mathbb{R}^3 : \|x\| < 1 \}$?

This $M$ has no boundary. We can consider it as a degenerate case of a manifold-with-boundary, but all of its charts map to an open subset of $\mathbb{R}^3$, never to an open set in $\mathbb{R}^2 \times [0,\infty)$ including the origin. $M$ is not compact and is actually diffeomorphic to the $\mathbb{R}^3$ manifold.