This just has to do with the charts $(U,\phi)$, and $(V,\psi)$ being "centered" at $p$ and $c$, i.e., $\phi(p)=\vec{0}\in \mathbb{R^m}$ and $\psi(c)=\vec{0}\in \mathbb{R}^n$.
By rank theorem, we have $\psi \circ \Phi \circ \phi^{-1}: (x^1,\ldots,x^r,x^{r+1},\ldots,x^m)\mapsto (x^1,\ldots,x^r,0,\ldots,0) $ as you said. And if we want a point $q\in M$ to map specifically to $c$, we should have
$$
\Phi(q)=c \iff \psi(\Phi(q))= (0,\ldots,0) \iff \psi\Phi\phi^{-1}(\phi(q)) = (0,\ldots,0) \iff \text{the first r coordinates of }\phi(q) \text{ are zero}
$$
Maybe this is not relevant for you anymore but i'll post some of my work here.
Denote $R=f^{-1}((\infty,b]) \subseteq M$. We need to show that $R$ is a regular domain, which is by definition $R\subseteq M$ is a smooth manifold with boundary with the inclusion map $i : R \hookrightarrow M$ is a smooth embedding and also $i : R \hookrightarrow M$ is a proper map.
It is easy to show that if $i : R \hookrightarrow M$ is a topological embedding, then the topology of $R$ is the subspace topology. So let's equip $R$ with its subspace topology. Next we will find smooth (boundary) charts for $R=f^{-1}(-\infty,b) \cup f^{-1}(\{b\})$. Let $B = f^{-1}(\{b\})$.
$\textbf{Interior Charts for } f^{-1}(-\infty,b) $ : This is easy since $f^{-1}(-\infty,b) = R\setminus B$ is open in $M$. So (as you may have notice), smooth charts for any $p \in R \setminus B$ comes from those charts in $M$ restricted to $R \setminus B$.
$\textbf{Boundary Charts for }f^{-1}(\{b\}) :$ Suppose $p \in B = f^{-1}(b)$. Since $p$ is a regular point then we can find neighbourhood $U$ of $p$ so that $f|_{U}$ is a local height function of $R$, that is $(U, \varphi)$ is a smooth chart of $M$ such that the representation of $f$ is $$\widehat{f}(x^1,\dots,x^n) = x^n .$$ Note that any $x \in \varphi(B \cap U)$, $\widehat{f}(x) = \widehat{f}(x^1,\dots,x^n)=x^n=b$. that is points of $B \cap U$ have the last coordinate $x^n=b$. Since we want these kind of points become boundary points of $f^{-1}(-\infty,b]$, we translate them using diffeomorphism $\psi : \widehat{U} \to \psi(\widehat{U})$, where $\varphi(U)=\widehat{U}$, defined as
$$\color{green}{
\psi(x^1,\dots,x^n) = (x^1,\dots,x^{n-1},b-x^n).}
$$
So we have new chart $(U,\phi)$ as $\phi :=\psi \circ \varphi : U \to \phi(U)$. Certainly $\phi(p) = (x_p^1,\dots,x_p^{n-1},0) \in \partial \mathbb{H}^n$ any other points in $f^{-1}(\infty,b) \cap U$ have positive last coordinates. This is our desired boundary chart for $p$.
So $R$ is a smooth $n$-manifold with boundary, and it is not hard to see that $i : R \hookrightarrow M$ is proper smooth embedding.
Best Answer
Up to the point you have highlighted, we have shown that for any point $p$ in $S$, we can define a chart $(\psi,V)$, where $V$ is open in $S$ and contains $p$, with $\psi\colon V\to \widehat V\subset \Bbb R^k$ a homeomorphism. The inclusion map $\iota\colon S\hookrightarrow M$ is a topological embedding because $S$ is given the subspace topology, hence $\iota$ is a homeomorphism onto its image.