Let $M$ be an oriented $k$-manifold in $\mathbb{R}^n,$ where $1\leq k \leq n.$ Let $\omega$ be a continuous $k$-form in $\mathbb{R}^n$, vanishing outside a compact set $K$. Suppose $\alpha$ and $\beta$ are local coordinates for $M$, and $K \subset \tilde{U_\alpha} \cap \tilde{U_\beta}.$ Then $\int_{{V_\alpha}}\alpha^*(\omega)=\int_{{V_\beta}}\beta^*(\omega)$.
This is lemma $14.9$ in Mathematical Analysis by Andrew Browder. To prove it, he states first that $\int_{{V_\beta}}\beta^*(\omega)=\int_{{V_{\alpha\beta}}}\beta^*(\omega),$ where $V_{\beta}=\beta^{-1}(U_\beta)$ and $V_{\alpha \beta} = \beta^{-1}(U_\alpha \cap U_\beta)$. My guess is this equality holds because of compact support, but I am unable to work out the details. Why does it hold?
Note that $$\beta^{-1}(K) \subseteq \beta^{-1}(\tilde{U}_\alpha \cap \tilde{U_\beta)} = \beta^{-1}(\tilde{U}_\alpha) \cap \beta^{-1}(\tilde{U}_\beta) \subseteq \beta^{-1}(\tilde{U}_\beta).$$ Suppose $\mathbf{p} \in \tilde{V}_{\beta}$. Note that $\mathbf{p} \notin \beta^{-1}(K)$ implies that $\beta(\mathbf{p}) \notin K$. Thus, $${(\beta^*\omega)}_{\mathbf{p}}(\mathbf{v}_1, \ldots, \mathbf{v}_r)=\omega_{\mathbf{\beta(p)}}(\mathbf{\beta'(p)v}_1, \ldots, \mathbf{\beta'(p)v}_r)=0,$$ since $\omega$ vanishes outside of $K.$ Therefore, $$\int_{{V_\beta={\beta^{-1}(\tilde{U}_\beta)}}}\beta^*(\omega)=\int_{{V_{\alpha\beta}={\beta^{-1}(\tilde{U}_\alpha \cap \tilde{U_\beta)}}}}\beta^*(\omega),$$ as the integral vanishes for all points $\mathbf{p}$ outside $\beta^{-1}(K)$!