In Bott&Tu's well-known book "Differential forms in Algebraic topology", they note -(p34): every form on $\mathbb{R}^n \times \mathbb{R}$ can be decomposed uniquely as a linear combination of two types of forms:
Type 1: $\pi^*(\phi) f(x,t)$
Type 2: $\pi^*(\phi) f(x,t) dt.$
Here $\phi$ is a form on $\mathbb{R}^n.$
Then (on p35) they add : -If $\{U_{\alpha}\}$ is an atlas for $M$ then $\{ U_{\alpha} \times \mathbb{R} \}$ is an atlas for $M \times \mathbb{R}.$ Again every form on $M \times \mathbb{R}$ is a linear combination of forms of type (1) and type (2).
My question: I agree with the claim on p34. I also agree that, given a form on $M \times \mathbb{R},$ I can split it canonically as $\text{ker}(i^*) \oplus (1- \text{ker}(i^*))$ (where $i: M\to M \times \mathbb{R}$ is the zero section). The two summands will then be $\textbf{locally}$ of the form claimed by Bott and Tu.
However, I don't understand why this decomposition should hold globally, as Bott&Tu seem to be arguing (they repeat the claim on p61).
Hint: Take a partition of the unity $(f_{\alpha})$ subordinate to $(U_{\alpha})$, and for each form $v$, let $v_{\alpha}$ be the restriction of $v$ to $U_{\alpha}$, write $v_{\alpha}=v^1_{\alpha}+v^2_{\alpha}$ where $v^1_{\alpha}$ is of type 1 and $v^2_{\alpha}$ is of type 2. Then write $v_1=\sum_{\alpha}f_{\alpha}v^1_{\alpha}$ and $v^2_{\alpha}=\sum_{\alpha}f_{\alpha}v^2_{\alpha}$.