I am currently trying to work through the proof of Stokes' Theorem given in Lee's Introduction to Smooth Manifolds (2. Edition). I've understood everything for the special cases of $\mathbb{H}^n$ (the upper half space) and $\mathbb{R}^n$, but I'm now stuck at the following step where we assume the (compact) support of an $(n-1)$-form $\omega$ is contained in a single smooth chart $(\varphi, U) \in \mathcal{A}$, where $\mathcal{A}$ is an atlas of the manifold $M$ (found on page 413).
In equation (16.6) the book says $$ \int_M d\omega = \int_{\mathbb{H}^n} (\varphi^{-1})^* d\omega, $$ which does follow the definition of integrals of forms on manifolds in terms of the pullback, however, I don't understand why $\varphi(U) \subseteq \mathbb{H}^n$. In other words: Why do we assume $\varphi : U \to \mathbb{H}^n$, instead of the more general $\varphi : U \to \mathbb{R}^n$?
A possible solution I've thought about: We can construct $\mathbb{R}^n$ with $\mathbb{R}^n = \mathbb{H}^n \cup (-\mathbb{H}^n)$, where I've defined $-\mathbb{H}^n := \{ x \in \mathbb{R}^n : x^n \leq 0 \}$. Thus, we can split the integral into two, i.e. $$ \int_M d\omega = \int_{\mathbb{H}^n \cup (-\mathbb{H}^n)} d\omega = \int_{\mathbb{H}^n} d\omega + \int_{-\mathbb{H}^n} d\omega. $$ Since $\mathbb{H}^n$ and $-\mathbb{H}^n$ can be obtained from one another with a diffeomorphism that essentially swaps the sign of the $x^n$ coordinate, we can focus on one only and then apply that diffeomorphism, automatically giving us the remaining half of $\mathbb{R}^n$. Does that solve the issue already? If not, please let me know what the correct solution to this issue is, and especially how I'm wrong.