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.
Let me answer your last question.
Lemma. Suppose that $X, Y, Z$ are Hausdorff topological spaces, $f:X\to Y, g: X\to Z$ are continuous and $f$ is proper. Then $F=f\times g: X\to Y\times Z$ is also proper.
Proof. Consider a compact $K\subset Y\times Z$; I'll prove that its preimage is compact. Let $A, B$ denote the
projections of $K$ to $Y, Z$ respectively; both are compact. Then $F^{-1}(K)\subset f^{-1}(A)\cap g^{-1}(B)$, a closed subset of the compact $f^{-1}(A)$. Hence $F^{-1}(K)$ is compact. qed
Best Answer
I found this proof confusing, too. Here is my understanding: we have that $D$ is an embedded submanifold of $M.$ Take any $p\in \partial D$. Then, there is a chart $\textit{in M},\ $ say, $(V,(x^1,\cdots,x^n))$ about $p$ such that $(D\cap V,(x^1,\cdots,x^k))$ is a boundary (slice) chart for $D$ about $p$. Thus, $q\in D\cap V\Rightarrow x^k(q)\ge 0$ But $\text{dim}\ D=\text{dim}\ M\Rightarrow k=n$, and so $x^n\ge 0.$ Now, $M$ is a manifold without boundary, which means that there must be a point $q\in V$ such that $x^n(q)<0$, (because $(V,(x^1,\cdots,x^n))$ is a chart about $p$ in $M$), which in turn implies that $q\notin D$ (because $D\cap V$ has all $x^n\ge 0$). Therefore, $V$ contains points in $D$ and in $M\setminus D$.