Splitting exterior derivatives on product manifolds Let $M_1 \times M_2$ be the product of two smooth manifolds $M_1, M_2$, and let
$\mu$ be a $k$-form on the product. Set $n_1 := \dim M_1$ and $n_2 := \dim M_2$.
The question is whether the product manifold structure induces a canonical splitting of the exterior derivative in the sense that
$$ \operatorname{d}\negthinspace \mu = \operatorname{d}_1 \negthinspace \mu + 
\operatorname{d}_2 \negthinspace \mu \, .$$
Indeed, if one looks at the coordinate expression
$$\mu = \frac{1}{k!} \, \mu_{i_1 \dots i_k} \, \operatorname{d}\negthinspace x^{i_1}\wedge 
\dots \wedge \operatorname{d}\negthinspace x^{i_k} \, ,$$
in product coordinates
$$\left(x^1, \dots, x^{n_1},x^{n_1+1}, \dots, x^{n_1 +n_2} \right) 
= \left(x^1_1, \dots, x^{n_1}_1,x^{1}_2, \dots, x^{n_2}_2 \right) \, , $$
then
\begin{align}
\operatorname{d}\negthinspace \mu 
&= \frac{1}{k!}  \, \left( \sum_{j_1=1}^{n_1}  \frac{\partial \mu_{i_1 \dots i_k}}{\partial x^{j_1}_1} \, 
\operatorname{d} \negthinspace x_1^{j_1}
 + \sum_{j_2=1}^{n_2} \frac{\partial \mu_{i_1 \dots i_k}}{\partial x^{j_2}_2} \, 
\operatorname{d} \negthinspace x_2^{j_2}  
\right)\wedge \operatorname{d}\negthinspace x^{i_1}\wedge 
\dots \wedge \operatorname{d}\negthinspace x^{i_k} 
\\
&=  \frac{1}{k!}  \, \left( \sum_{j_1=1}^{n_1}  \frac{\partial \mu_{i_1 \dots i_k}}{\partial x^{j_1}_1} \, 
\operatorname{d} \negthinspace x_1^{j_1} 
\right) \wedge \operatorname{d}\negthinspace x^{i_1}\wedge 
\dots \wedge \operatorname{d}\negthinspace x^{i_k}   
 +  \frac{1}{k!}  \, 
\left( \sum_{j_2=1}^{n_2} \frac{\partial \mu_{i_1 \dots i_k}}{\partial x^{j_2}_2} \, 
\operatorname{d} \negthinspace x_2^{j_2}  
\right)\wedge \operatorname{d}\negthinspace x^{i_1}\wedge 
\dots \wedge \operatorname{d}\negthinspace x^{i_k}  \, .
\end{align}
(Of course, one needs to anti-symmetrize to get the components of the $(k+1)$-forms.)
Is there a way to define this invariantly e.g. by using projections onto some subbundles of
the tangent bundle/tensor bundles of the product? A quick search in the literature did not really yield anything. One would expect this to be more common, if this were a thing.
EDIT: To maybe give some motivation/context for why one would want to do this, consider the projection mappings onto the respective factor
$$\operatorname{pr}_i \colon M_1 \times M_2 \to M_i \colon (m_1,m_2) \mapsto m_i \, ,$$
a differential form $\alpha$ on say $M_1$, as well as a vector field $X$ on the product $M_1 \times M_2$. One shows easily that formally
$$X \lrcorner \left( \operatorname{pr}_1^* \alpha \right) = \operatorname{pr}_1^* \Bigl( \bigl((\operatorname{pr}_1)_* X \bigr) \lrcorner \alpha \Bigr)  \, .$$
($\lrcorner$ is contraction, the upper and lower stars denote pullback and pushforward, respectively)
However, the expression in parentheses on the right hand side is not a differential form on $M_1$, so
$$\operatorname{d} \left( \negthinspace \operatorname{pr}_1^* \Bigl( \bigl((\operatorname{pr}_1)_* X \bigr) \lrcorner \alpha \Bigr) \right)
\neq 
\operatorname{pr}_1^* \left( \operatorname{d}   \Bigl( \bigl((\operatorname{pr}_1)_* X \bigr) \lrcorner \alpha \Bigr) \right)$$
It is the splitting I was asking for that would allow this equality to hold, if $\operatorname{d}$ were replaced by $\operatorname{d} \negthinspace _1$ on the left hand side.
 A: Let $\pi_1,\pi_2$ denote the canonical projections and $A_1=\pi_1^*TM_1$, $A_2=\pi_2^*TM_2$. We have the canonical decomposition of vector bundles which splits $T(M_1\times M_2)$ into "horizontal" and "vertical" parts.
$$
T(M_1\times M_2)\cong A_1\oplus A_2
$$
This in turn gives rise to a decomposition of exterior powers
$$
\Lambda^kT^*(M_1\times M_2)\cong\bigoplus_{p+q=k}\Lambda^pA_1^*\otimes\Lambda^qA_2^*
$$
which gives us projections $P^{p,q}$ to each factor of this direct sum. We can use these to write the decomposition as
$$
d_1=\sum_{p+q=k}P^{p+1,q}\circ d\circ P^{p,q},\ \ \ \ \ d_2=\sum_{p+q=k}P^{p,q+1}\circ d\circ P^{p,q}
$$
That said, this decomposition may not do what you want it to. The equality you give:
$$
X\lrcorner\pi_1^*\alpha=\pi_1^*(\pi_{1*}X\lrcorner\alpha)
$$
Doesn't make sense, since the pushforward $\pi_{1*}X$ isn't generally well defined. In fact, $X\lrcorner\pi_1^*\alpha$ cannot generally be expressed as a pullback. One can write a similar expression using Cartan's magic formula, but an extra Lie derivative appears:
$$
d(X\lrcorner\pi_1^*\alpha)=\mathcal{L}_X(\pi_1^*\alpha)-X\lrcorner\pi_1^*(d\alpha)
$$
