Differential forms on cartesian product Let $M,N$ be smooth manifolds.
Is it true, that any $k$-form $\alpha$ on $M\times N$ is of the form $$\alpha = \sum_{p+q=k} p_M^* \beta_p \wedge p_N^* \gamma_q,$$
where $p_M, p_N$ are the projections and $\beta_p$ is a $p$-form on $M$ and $\gamma_q$ is a $q$-form on $N$?
I know that it is true on the level of cohomology. 
 A: There is a subtle point involved in the question, namely:
What is not true
It is not true (as noted by s.harp in the comments) that every $k$-form $\alpha_k$ on $M\times N$ can be written $$\alpha = \sum_{p+q=k} p_M^* \beta_p \wedge p_N^* \gamma_q,$$ 
For example on $\mathbb R^m\times \mathbb R^n$ the $0$-form $\alpha_0=\sin(x_1\cdots x_m y_1\cdots y_n)$ cannot be written as $\alpha_0=f(x_1,\cdots, x_m)\cdot g(y_1,\cdots ,y_n)$
 And the $m+n$-form $\alpha_{m+n } =\sin(x_1\cdots x_m y_1\cdots y_n)dx_1\cdots dx_mdy_1\cdots dy_n$ cannot be written as $\alpha_{m+n } =f(x_1,\cdots, x_m)dx_1\cdots dx_m \wedge g(y_1,\cdots ,y_n)dy_1\cdots dy_n$
What is true
If we define the vector bundles $E^p:=\pi^*_M(\bigwedge^p T^*_M), F^q:=\pi^*_N(\bigwedge^q T^*_N)\; $ on $M\times N$, then we have an isomorphism of vector bundles on $M\times N$ $$\bigwedge^kT^*_{M\times N}=\oplus_{p+q=k}(E^p \otimes   F^q) $$ Taking global sections we get the correct equality   $$\Gamma(M\times N,\bigwedge^{k} T^*_{M\times N})=\oplus_{p+q=k}[\Gamma(M\times N,E^p)\otimes_{C^\infty (M\times N)} \Gamma(M\times N, F^q)]$$ a canonical isomorphism of $C^\infty$-modules.  
Here I have used the non-trivial fact that given two vector bundles $V,W$ on the manifold $X$ we have $\Gamma(X,V\otimes W)=\Gamma(X,V)  \otimes_{C^\infty (X)}     \Gamma(X,V)
$.
Notice that the tensor product is over $C^\infty (X)$, not $\mathbb R$.
A proof can be found for example in Conlon's Differentiable Manifolds, page 232]
Why there is no contradiction between the two claims above
Because we can write in the example above  $$\alpha_{m+n } =\sin(x_1\cdots x_m y_1\cdots y_n)dx_1\cdots dx_m dy_1\cdots dy_n=\\
\sin(x_1\cdots x_m y_1\cdots y_n)dx_1\cdots dx_m \wedge 1\cdot dy_1\cdots dy_n$$ and the crucial point is that $$\sin(x_1\cdots x_m y_1\cdots y_n)dx_1\cdots dx_m \in \Gamma(\mathbb R^m\times \mathbb  R^n,E^m)=\Gamma(\mathbb R^m\times \mathbb  R^n ,\pi^*_{\mathbb R^m}(\bigwedge^m T^*_{\mathbb R^m}))$$
