Decomposition of forms in the proof of Poincaré's lemma On page 34 of Bott and Tu, Differential Forms in Algebraic Topology, where the author proves 
that the maps  $\pi: \Bbb R ^n \times \Bbb R^1 \rightarrow \Bbb R^n$, $s: \Bbb R^n \rightarrow \Bbb R^n \times \Bbb R^1$ ($x \mapsto (x,0))$  induce an isomorphism 
    $$H^*(\Bbb R^n \times \Bbb R^1) \rightarrow H^*(\Bbb R^n)$$
in the de Rham cohomology, he stated that 

Every form on $\Bbb R^n \times \Bbb R$ is uniquely a linear combination of the following two types of forms: 
  
  
*
  
*$(\pi^* \phi ) f(x,t)$,
  
*$(\pi^* \phi) f(x,t) dt$,
  
  
  where $\phi$ is a form on the base $\Bbb R^n$. 

The author states the same argument works when $\Bbb R^n$ is in general replaced by a manifold $M$. But the argument here requires that we work locally on charts $\{U_\alpha \times \Bbb R^1 \}$? 
I could not really make sense of how to write this "coordinate independent" argument out rigorously: 
I know we can take a partition of unity such that all forms are of linear combination of given form. But I do not make sense of "uniqueness". 
 A: Not entirely sure if this is what you're stuck on, but:
More explicitly, uniqueness in the general case means that given any form $\omega \in \Omega^q(M \times \mathbb R) \subset \Omega^*(M \times \mathbb R)$, there exists a unique decomposition $\omega = \omega_1 + \omega_2$ where $\omega_1 = (\pi_1^* \phi_1)f_1$ and $\omega_2 = (\pi_1^* \phi_2)(\pi_2^* dt)f_2$ for some $\phi_1 \in \Omega^q(M \times \mathbb R)$, $\phi_2 \in \Omega^{q-1}(M \times \mathbb R)$, and $f_1, f_2 \in C^\infty(M \times \mathbb R)$. Here $\pi_1$ and $\pi_2$ are the projections onto the respective factors. Note that the $\phi_i$ and the $f_i$ themselves are not necessarily unique.
An important point for the question is that the form $\pi_2^* dt$ (or just $dt$) exists uniquely, independent of the coordinate (up to some coefficient which could be absorbed into the factor $(\pi_1^* \phi_2) f_2$ instead) due to the product structure. It's characterized by the property that its null space at each point is the embedded corresponding tangent space of $M$; add any form of the type $(\pi_1^* \phi) f \neq 0$ to it and we lose that property. And it's natural to give special treatment to $dt$ in the proof: The extra factor of $\mathbb R$ is the difference between the two base manifolds, after all.
