The curvature tensor $R$ of a connection $\nabla$ on a vector bundle $E \to M$ is a 2-form with values in the bundle $\mathrm{End}(E)$; i.e. $R \in \Omega^{2}(\mathrm{End}(E))$. It acts on a pair of vector fields $X,Y$ by the formula
$$ R(X,Y) = \nabla_{X}\nabla_{Y} - \nabla_{Y}\nabla_{X} - \nabla_{[X,Y]} $$ which is a map $\Gamma(E) \to \Gamma(E)$. I'm wondering if this formula is a special case of a more general notion of an $E$-valued $k$-form acting on $k$ vector fields $X_{1},\ldots, X_{k}$ to produce a map $\Gamma(E) \to \Gamma(E)$ Of course, this would generalize the idea of an element $\omega \in \Omega^{k}(M)$ acting on vector fields according to the usual formula $$ \omega(X_{1},\ldots,X_{k}) = \sum_{i_{1},\ldots,i_{k}}\omega_{i_{1}\cdots i_{k}}X_{1}^{i_{1}}\cdots X_{k}^{i_{k}} \in C^{\infty}(M) $$ To state the question more explicitly: for simplicity, say $\alpha \in \Omega^{2}(E) = \Gamma(E)\otimes\Omega^{2}(M)$. Given vector fields $X$ and $Y$, is there a natural way to define the action $\alpha(X,Y)$ in a way that generalizes the other two actions?
There are several aspects here: Initially, an $E$-valued $k$-form is defined as associating to each $x\in M$ a $k$-linear, alternating map $(T_xM)^k\to E_x$, where $E_x$ is the fiber over $x$. In this language, you can deal perfectly with the case that $\alpha=\omega\otimes s$, for which this is just the map $X,Y\mapsto \omega(x)(X,Y)s(x)$. As for ordinary forms, by considering $x$ as a variable, you can view an $E$-valued $k$-form as defining an operator $(\mathfrak X(M))^k\to\Gamma(E)$. Still similar to standard forms, one proves that such an operator comes from an $E$-valued $k$-form (i.e. is point-wise or tensorial) if and only if it is linear over smooth functions in one or equivalently any variable.
To deal with the curvature, you first have to observe that sections of $End(E)$ are equivalent to vector bundle homomorphisms $E\to E$ with base map $id_M$. These in turn are equivalent to linear operators $\Gamma(E)\to\Gamma(E)$ that are linear over smooth functions (see the comment of @jawheele ). To interpret the definition of curvature, you first have to observe that the defining equation $\nabla_X\nabla_Y s-\nabla_Y\nabla_X s-\nabla_{[X,Y]}s$ is linear over smooth functions in $s$. This shows that you can view it as associating to two vector fields $X$ and $Y$ an operator $\Gamma(E)\to\Gamma(E)$ that is linear over smooth functions and thus a section of $End(E)$. Then you verify that this map is linear over smooth functions in $X$ and $Y$ and then conclude that it defines an $End(E)$-valued $2$-form.
By the way, $\Omega^2(E)$ is not isomorphic to $\Gamma(E)\otimes\Omega^2(M)$ since for a smooth function $f$, $s\otimes(f\omega)$ and $(fs)\otimes\omega$ define the same $E$-valued form. You can write is as a tensor product over $C^\infty (M,\mathbb R)$, though.