Splitting of the tangent bundle of a vector bundle and connections

The covariant derivative $\nabla$ is a map $\nabla: \Gamma(TM)\times \Gamma(E)\longrightarrow \Gamma(E)$ which is $\mathbb R$-bilinear and satisfies $$\nabla_{f\cdot X} s=f\cdot \nabla_X s\quad \textrm{and}\quad \nabla_\alpha (f\cdot s)=f\cdot \nabla_X s+\mathcal{L}_{X}(f)\cdot s,$$ for every $f\in C^\infty(M)$, $X\in\Gamma(TM)$ and $s\in \Gamma(E)$. We can then define $$d_\nabla: \Gamma(\Lambda^p T^*M\otimes E)\longrightarrow \Gamma(\Lambda^{p+1} T^*M\otimes E),$$ setting $$\begin{eqnarray*} d_\nabla \varepsilon(X_1, \ldots, X_{p+1})&&:=\sum_{j=1}^{p+1}(-1)^{j+1} \nabla_{X_j} \varepsilon(X_1, \ldots, \widehat{X_j}, \ldots, X_j)\\ &&+\sum_{i<j} (-1)^{i+j} \varepsilon([X_i, X_j], X_1, \ldots, \widehat{X_i}, \ldots, \widehat{X_j}, \ldots, X_{p+1}).\end{eqnarray*}$$ To define this I'm using the isomorphism:

$$\Gamma(\Lambda^p T^* M\otimes E)\simeq \mathsf{Hom}_{C^\infty(M)} (\Lambda^p \Gamma(TM), \Gamma(E))\simeq \mathsf{Alt}^p_{C^\infty(M)}(\Gamma(TM), \Gamma(E)).$$

If I'm not wrong, $d_\nabla^2=0$ if and only if $\nabla$ is a flat connection.

Up to this point everything is ok. Now, my guess is that you have a product:

$$\cdot:\Gamma(\Lambda^p T^*M)\times \Gamma(\Lambda^q T^* M\otimes E)\longrightarrow \Gamma(\Lambda^{p+q} T^*M \otimes E), $$ defined in the obvious way and it holds:

$$d_\nabla (\alpha\cdot \omega)=d_{\mathsf{dR}}\alpha\cdot \omega+(-1)^{|\alpha|} \alpha \cdot d_\nabla \omega. $$

Hope it helps.

P.s: If you want to know where those things came from look for Lie algebroid representations. If I'm not wrong you can find something here.