Let $E\rightarrow M$ be an orientable rank $n$ real vector bundle over a smooth manifold $M$. Bott and Tu define integration along the fibre as follows:
Let $\phi_\alpha$ be a trivialization $E|_{U_\alpha}\rightarrow U_\alpha\times \mathbb{R}^n$, and suppose that we have coordinates $x^i$ on $U_\alpha$, and $t^1,\dots, t^n$ on $\mathbb{R}^n$. Then every differential $k$ form on $E$ with vertical compact support can be written locally on $\pi^{-1}(U_\alpha)$ as linear combinations of forms of the type: \begin{align*} \omega=\pi^*(\phi)f(x,t)dt^{i_1}\wedge \cdots \wedge dt^{i_r} \end{align*} and: \begin{align*} \omega=\pi^*(\phi)f(x,t)dt^{1}\wedge \cdots \wedge dt^{n} \end{align*} where $\phi$ is a form on $U_\alpha$ and $f(x,t)$ has compact support for each fixed $x$. Integration along the fibre is then defined on forms of the first type by sending them to zero, and forms on the second type by: \begin{align*} \pi_*(\omega)\longmapsto \phi\int_{\mathbb{R}^n}f(x,t)dt^1\cdots dt^n \end{align*}
I am trying to prove that if $\phi_\alpha$ and $\phi_\beta$ are two separate trivializations, that the second definition is well defined. That is if: \begin{align*} \omega_\alpha:=\omega|_{\pi^{-1}(U_\alpha)}=\pi^*(\phi)f(x,t)dt^1\wedge \cdots \wedge dt^n \end{align*} and: \begin{align*} \omega_\beta:=\omega|_{\beta^{-1}(U_\alpha)}=\pi^*(\tau)g(y,s)ds^1\wedge \cdots \wedge ds^n \end{align*} then: \begin{align*} \pi_*(\omega_\alpha)=\pi_*(\omega_\beta) \end{align*} on $U_\alpha\cap U_\beta$.
I know that the since $\omega$ is globally defined on $E$, that we must have that: \begin{align*} \omega_{\alpha}|_{\pi^{-1}(U_\alpha\cap U_\beta)}=\omega_{\beta}|_{\pi^{-1}(U_\alpha\cap U_\beta)} \end{align*} So I want to write that: \begin{align*} \pi_*(\omega_\alpha)=&\phi \int_{\mathbb{R}^n}f(x,t)dt^1\wedge \cdots dt^n\\ =& \int_{\mathbb{R}^n}\pi^*(\phi)f(x,t)dt^1\wedge \cdots dt^n\\ =&\int_{\mathbb{R}^n}\pi^*(\tau)g(y,s)ds^1\wedge \cdots ds^n\\ =&\tau\int_{\mathbb{R}^n}g(y,s)ds^1\wedge \cdots ds^n \end{align*} Where, the restrictions have been supressed, and taking the integral: \begin{align*} \int_{\mathbb{R}^n}\pi^*(\phi)f(x,t)dt^1\wedge \cdots dt^n \end{align*} is abuse of notation for saying decompose $\phi$ into some linear combination of $dx^{i_1}\wedge \cdots dx^{i_{k-n}}$'s, and integrate the pullback of the coefficients, which would be constant along the fibres. However, this feels really sloppy, and I feel like there should be a better way of doing which makes use of transition functions, but I keep getting slogged down in notation.
Does anyone have a better way of doing this? I have seen similar questions on this site, namely here, but I don't understand why they integrate across $U_\alpha \cap U_\beta$ as well as $\mathbb{R}^n$. Furthermore, they take the case $k-n=\dim M$ without loss of generality, which doesn't make much sense to me at all. Any help would be greatly appreciated.
Edit: I think to make my argument above more precise, we can say some thing as follows:
The map $\pi^*:\Omega^*(M)\rightarrow \Omega^*(E)$ is injective, so examine the forms: \begin{align*} \pi_*(\omega_\alpha)=&\phi\int_\mathbb{R^n}f(x,t)dt^1\wedge \cdots \wedge dt^n \end{align*} and: \begin{align*} \pi_*(\omega_\beta)=\tau\int_\mathbb{R^n}g(y,s)ds^1\wedge \cdots \wedge ds^n \end{align*} In particular, the integrals are smooth function on the open sets $U_\alpha$ and $U_\beta$, while $\phi$ and $\tau$ are a smooth collection of forms on the same sets. If they pull back to the same elements in $\Omega^*(E_{U_\alpha\cap U_\beta}) $it follows then that they are equal on the $\Omega(U_\alpha\cap U_\beta)$.
In coordinates the pull back of the of the functions defined by the integral is the same function, and the forms pull back to forms which are constant on the fibres. WLOG, assume $\phi$ and $\tau$ are of degree $m$, we can rewrite $\phi$ and $\tau$ as: \begin{align*} \phi=&\phi_{i_1\cdots i_m}dx^{i_1}\wedge \cdots \wedge dx^{i_m}\\ \tau=&\tau_{i_1\cdots i_m}dy^{i_1}\wedge \cdots \wedge dy^{i_m} \end{align*} We then see that on the overlap: \begin{align*} \omega_{\alpha}=&(\phi_{i_1\cdots i_m}\circ \pi)d(x^{i_1}\circ \pi)\wedge \cdots \wedge d(x^{i_m}\circ \pi)\wedge (f(x,t)dt^1\wedge \cdots dt^n)\\ =&(\tau_{i_1\cdots i_m}\circ \pi)d(y^{i_1}\circ \pi)\wedge \cdots \wedge d(y^{i_m}\circ \pi)\wedge(g(y,s)ds^1\wedge \cdots \wedge ds^n)\\ =&\omega_\beta \end{align*} It follows that: \begin{align*} (\phi_{i_1\cdots i_m}\circ \pi)d(x^{i_1}\circ \pi)\wedge \cdots &\wedge d(x^{i_m}\circ \pi)\int_{\mathbb{R}^n}(f(x,t)dt^1\wedge \cdots dt^n)\\ =&(\tau_{i_1\cdots i_m}\circ \pi)d(y^{i_1}\circ \pi)\wedge \cdots \wedge d(y^{i_m}\circ \pi)\int_{\mathbb{R}^n}(g(y,s)ds^1\wedge \cdots \wedge ds^n) \end{align*} as the functions $\tau_{i_1\cdots i_m}$ and $\phi_{i_1\cdots i_m}$ are constant on the fibres. The above then simplifies to: \begin{align*} \pi^*(\phi)\int_{\mathbb{R}^n}f(x,t)dt^1\cdots dt^n=\pi^*(\tau)\int_{\mathbb{R}^n}g(y,s)ds^1\cdots ds^n \end{align*} which from our earlier discussion on the integral functions gives us (again on the overlap): \begin{align*} \pi^*(\pi_*(\omega_\alpha))=\pi^*(\pi_*(\omega_\beta)) \end{align*} hence on $U_\alpha\cap U_\beta$: \begin{align*} \pi_*(\omega_\alpha)=\pi_*(\omega_\beta) \end{align*} so $\pi_*$ is a well defined map.
Does that make more sense?
Edit:
Following @user8268 we do the following:
Let $\omega\in \Omega^n(E)$, with $n\geq f$, and let $\{U_i,\psi_i\}$ be a cover of $M$ which trivializes $E$ such that each $U_i$ is a coordinate $(U_i,\phi_i)$. Since $E$ is locally diffeomorphic to the product $U\times F$, we have that each $\psi_i$ is a map: \begin{align*} \psi_i:E_{U_i}\longrightarrow U_i\times F \end{align*} We can then construct a coordinate cover for $E$ by taking a coordinate cover $(Y_j,\theta_j)$ for $F$ and defining charts by: \begin{align*} \phi_{ij}=(\phi_i\times \theta_j)\circ \psi_i: \psi^{-1}_i(U_i\times Y_j)\longrightarrow \phi_i(U_i)\times \theta_j(Y_j) \end{align*} Going forward we denote $\psi^{-1}_i(U_i\times Y_j)$ by $V_{ij}$. Note that we can now write $\omega$ in coordinates $(x^1,\dots, x^m, y^1,\dots, y^f)$ as a linear combination of $n$ forms of the type: \begin{align} \alpha=f(x,y)(\pi\circ \phi_{ij}^{-1})^*(\xi)\wedge dy^{j_1}\wedge \cdots \wedge dy^{j_k} \end{align} where $\xi\in\Omega^{n-k}(U_i)$, and of the type: \begin{align*} \eta=f(x,y)(\pi\circ \phi_{ij}^{-1})^*(\xi)\wedge dy^{1}\wedge \cdots \wedge dy^f \end{align*} We define $\pi_*$ on forms of the first type to be identically $0$, while on forms of the second type as: We define $\pi_*$ on forms of the first type to be identically $0$, while on forms of the second type as: \begin{align*} \pi_*(\eta)=\xi\int_{\theta_j(Y_j)} f(x,y)dy^1\wedge \cdots \wedge dy^n \end{align*} but from here I am unsure of how to use the partition of unity...any help would be appreciated.