Bott Tu Exercise 6.14, integration along the fiber

Suppose $$\pi:E\to M$$ is an oriented $$C^\infty$$ vector bundle of rank $$n$$. We denote by $$\Omega_{cv}^k(E)$$ the set of all differential $$k$$-forms $$\omega$$ on $$E$$, such that for each compact $$K\subset M$$, $$\pi^{-1}(K)\cap \text{supp}(\omega)$$ is compact. In particular, the support of the restriction $$\omega|_F$$ to each fiber is compact. Assume $$\{(U_\alpha,\phi_\alpha)\}$$ is an oriented trivialization for $$E$$. On $$\pi^{-1}(U_\alpha)$$, such a form $$\omega$$ is uniquely expreesed as a sum of the forms either of type $$(\pi^*\phi)f(x,t_1,\dots,t_n)dt_{i_1}\cdots dt_{i_r}$$ with $$r, or $$(\pi^*\phi)f(x,t_1,\dots,t_n)dt_1\cdots dt_n$$. (Here $$x_1,\dots,x_n$$ are coordinate functions on $$U_\alpha$$ and $$t_1,\dots,t_n$$ are fiber coordinates on $$\pi^{-1}(U_\alpha)$$ given by $$\phi_\alpha$$. We define a map $$\Omega_{cv}^*(E)\to \Omega^*(M)$$ by sending the forms of first type to zero, and the forms of second type to $$\phi \int_{\Bbb R^n} f(x,t_1,\dots,t_n)dt_1 \dots dt_n$$.

Exercise 6.14 askes to show that this map is well-defined. Suppose $$U_\alpha \cap U_\beta$$ is nonempty. Then on $$\pi^{-1}(U_\alpha \cap U_\beta)$$, a form of second type can be expressed as $$(\pi^*\phi)f(x,t_1,\dots,t_n)dt_1\cdots dt_n=(\pi^* \tau)g(y,u_1,\dots,u_n) du_1\cdots du_n.$$ Then I have to show that $$\phi \int_{\Bbb R^n} f(x,t_1,\dots,t_n)dt=\tau \int_{\Bbb R^n} g(y,u_1,\dots,u_n)du,$$ but I got stuck. Any hints?

Assume that $$U_a$$ and $$U_b$$ are relatively compact and $$\varphi$$, $$\psi$$ its charts on $$M$$. By a linearity argument, you can assume that

$$\phi = h_1 dx_{i_1} \wedge \cdots dx_{i_k}$$

and the same for $$\tau$$ with coefficient $$h_2$$. Without loss of generality assume $$k = m = \dim M$$. Let $$T$$ and $$S$$ the coordinates for $$t$$ and $$s$$ in $$\mathbb{R}^n$$. Take the adaptated chats in $$E$$ $$(U_a \times \mathbb{R}^n,\varphi \times T)$$ and $$(U_b \times \mathbb{R}^n,\psi \times S)$$.

We have $$\pi^*\phi = h_1 \circ \pi \pi^*(dx) = h_1dx$$ in these coordinates. So

$$\pi^* \phi f(x,t)dt = h_1(x)f(x,t)dx\wedge dt$$ On the other hand, using the change of variables for diff. forms $$\pi^* \tau gds = (h_2 \circ \psi \circ \varphi^{-1})(g \circ \psi \circ \varphi^{-1}) \cdot Jac(\psi \circ \varphi^{-1}) Jac(S \circ T^{-1}) dx \wedge dt = h_1 fdx\wedge dt$$ We deduce that $$h_1f = (h_2 \circ \psi \circ \varphi^{-1})(g \circ \psi \circ \varphi^{-1}) \cdot Jac(\psi \circ \varphi^{-1}) Jac(S \circ T^{-1})$$

The last expression can be integrated into $$(U_c = U_a \cap U_b) \times \mathbb{R}^n$$ and using Fubini and the change of variable for integrals and the previous definitions $$\int_{U_c \times \mathbb{R}^n}\phi f(x,t)dx\wedge dt = \int_{U_c \times \mathbb{R}^n}\tau g(y,s)dy \wedge ds.$$ Finally, by a parametrical argument and using $$U_a$$ and $$U_b$$ be compact

$$\phi \int_{\mathbb{R}^n} f(x,t)dt = \tau \int_{\mathbb{R}^n}g(y,s)ds.$$

You must be a little more cautious for $$k < n$$ because the change of variable for differential forms is a bit different. If I have some typo please fix it.