Consider the more general case where $\int_S f = \int_Q f_S$ exists but $f$ is not everywhere continuous in $Q$. Let $D_f\subset S$ denote the set of discontinuity points of $f$ in $S$.
With $A = int(S)$, the set $D$ of discontinuity points for $f_S$ is
$$D = (A \cap D_f) \,\cup\, (\partial S \cap D_f) \,\cup\, \{x_0 \in \partial S \setminus D_f: \lim_{x \to x_o, x \in A} f(x) \neq 0\} $$
Since $f_S$ is integrable on $Q$ the set $D$ is of measure zero.
On the other hand, $f_A$ vanishes everywhere on $\partial S$ and $f_A = f_S$ for all $x \in A$. Hence, the set $E$ of discontinuity points for $f_A$ is
$$E = (A \cap D_f) \,\cup \, \{x_0 \in \partial S: \lim_{x \to x_o, x \in A} f(x) \neq 0\}$$
Note that
$$ \{x_0 \in \partial S: \lim_{x \to x_o, x \in A} f(x) \neq 0\} \subset (\partial S \cap D_f) \,\cup\, \{x_0 \in \partial S \setminus D_f: \lim_{x \to x_o, x \in A} f(x) \neq 0\} $$
Hence, $E\subset D$ and $E$ is also of measure zero.
The argument that $\int_S f = \int_A f$ remains the same. It is irrelevant if $f_S - f_A$ does not vanish at points of $D\setminus E$ since $D\setminus E \subset D$ is of measure zero.
There is definitely something to show: A priori it might be that for a given point $p\in M$ there exist two different charts - one that takes $p$ to the boundary of $\mathbb{H}^k=[0,\infty)\times \mathbb{R}^{k-1}$ and one that takes it to the interior. We have to show that this is impossible.
Suppose we had such charts, denoted with $\varphi_i:U\rightarrow \mathbb{H}^k$ ($i=1,2$) for which $\varphi_1(p)\in \mathrm{int}\mathbb{H}^k$ and $\varphi_2(p)\in \partial\mathbb{H}^k$. Then there would be a small neighbourhood $V\subset \mathrm{int}\mathbb{H}^k$ around $x=\varphi_1(p)$ such that the composition $f = \varphi_2\circ\varphi_1^{-1}\vert_V:V\rightarrow f(V)\subset\mathbb{H}^k$ was a diffeomorphism onto its image. Then $f(x)\in \partial \mathbb{H}^k$, which is forbidden by the following Lemma:
Lemma. Suppose $V\subset \mathrm{int}\mathbb{H}^k$ is open and $f:V\rightarrow f(V)\subset \mathbb{H}^k$ is a diffeomorphism onto its image. Then $f(V)\cap \partial \mathbb{H}^d = \emptyset$.
This Lemma, depending on your point of departure, is definitely non-trivial. E.g. in Lee's smooth manifold book it is Excercise 7.7 and follows from the earlier Proposition 7.16, which asserts that $f$, viewed as map into $\mathbb{R}^k$, is open. But $f(V)$ is only an open subset of $\mathbb{R}^k$ if it does not intersect $\partial \mathbb{H}^k$.
Another way of proving the Lemma is using algebraic topology - and this is what I alluded to in my comment about 'poking holes'. This might be a somehwat heavy gun for the situation at hand, but has the advantage that one can relax the requirement from $f$ being a diffeomorphism to it merely being a homeomorphism. The idea is as follows: Without loss of generality you can assume that $V$ is an open ball around $x$, but then $f\vert_{V\backslash x}:V\backslash x \rightarrow f(V)\backslash f(x)$ is a homeomorphism between a space which is homotopic to $S^{k-1}$ and a space which is contractible. But this is impossible, as one can check e.g. by looking at homology groups.
Best Answer
As you suggest, there appears to be a typo. In the original definition of measure zero set on a manifold, the definition of $D_i$ should read $$D_i=\alpha_i^{-1}(D\cap V_i).$$ I am not aware of any errata to this book, but given the fact that there is no function $\alpha$ referenced elsewhere in this definition is very strong evidence that this is indeed a mistake, and that Munkres meant for $\alpha$ to be $\alpha_i$.