An exact one-form has zero period on every closed piecewise-C¹ curve

Fix the finite subdivision from [F1]. On each chartwise $C^1$ piece choose a chart $\kappa :U\to V\subseteq {\mathbb{R}}^n$ containing its image, and put $x(t):=\kappa (\gamma (t))$ and $\tilde{\varphi }:=\varphi \circ {\kappa }^{-1}$. By [F2] the coordinate coefficients of $\omega$ are ${\mathrm{\partial }}_i\tilde{\varphi }$; applying [F3] to $\tilde{\varphi }\circ x$ gives $\frac{d}{dt}(\varphi \circ \gamma )(t)={\omega }_{\gamma (t)}(\dot{\gamma }(t))$ in the piece interior. The same formula holds with one-sided derivatives at the endpoints: the differentiability expansion of $\tilde{\varphi }$ composed with the one-sided expansion of $x$ gives $D\tilde{\varphi }(x(t))x^{\mathrm{\prime }}(t)$. These derivatives are continuous on the closed piece by [F1] and smoothness of $\varphi$, so they are integrable by [F4].

Apply [F5] on each subdivision interval to $h=\varphi \circ \gamma$. By [F1] the period is the sum of those integrals, so it equals the sum of the increments $\varphi (\gamma (t_{j+1}))-\varphi (\gamma (t_j))$. The internal endpoint terms cancel, and the remaining increment is $\varphi (\gamma (1))-\varphi (\gamma (0))=0$ because $\gamma$ is closed. Thus the period is zero.