Curvature of a Vector Bundle as a 2-form Let $(E,\mathcal{M},\pi)$ be a smooth $\mathbb{R}$-vector bundle over some smooth manifold $\mathcal{M}$ with or without boundary.Furthermore, let
$$\nabla:\mathfrak{X}(\mathcal{M})\times\Gamma^{\infty}(\mathcal{M},E)\to \Gamma^{\infty}(\mathcal{M},E)$$
denote an arbitrary connection on $(E,\mathcal{M},\pi)$. Then the curvature of the bundle $E$ with respect to the connection $\nabla$ is a map $$F_{\nabla}:\mathfrak{X}(\mathcal{M})\times \mathfrak{X}(\mathcal{M})\times \Gamma^{\infty}(\mathcal{M},E)\to \Gamma^{\infty}(\mathcal{M},E)$$ defined as
$$F_{\nabla}(X,Y,s):=\nabla_{X}\nabla_{Y}s-\nabla_{Y}\nabla_{X}s-\nabla_{[X,Y]}s$$
Now, my goal is to proove that $F_{\nabla}\in\Omega^{2}(\mathcal{M},\mathrm{End}(E))$, where $\mathrm{End}(E)$ denotes the endomorphism bundle of $E$, i.e. the bundle with fibres given by $\mathrm{End}(E_{p})$. First of all, we can use that there is a $C^{\infty}(\mathcal{M})$-module isomorphism $$\Omega^{2}(\mathcal{M},\mathrm{End}(E))\cong \mathrm{L}_{\mathrm{alt}}^{2}(\mathfrak{X}(\mathcal{M}),\Gamma^{\infty}(\mathcal{M},\mathrm{End}(E)))$$
Furthermore, it is easy to prove that the map
$$F_{\nabla}:\mathfrak{X}(\mathcal{M})\times \mathfrak{X}(\mathcal{M})\to\Gamma^{\infty}(\mathcal{M},E), (X,Y)\mapsto F(X,Y,s)$$
for some fixed $s\in\Gamma^{\infty}(\mathcal{M},E)$ is $C^{\infty}(\mathcal{M})$-bilinear and antisymmetric. But how to continue? In the end we need to proof that
$$F(X,Y)\in \Gamma^{\infty}(\mathcal{M},\mathrm{End}(E))$$
How is this map defined and how to show smoothness?
 A: What is missing is the $C^{\infty}(\mathcal{M})$-module isomorphism $\Gamma^{\infty}(\mathcal{M},\mathrm{End}(E))\cong \mathrm{End}_{C^{\infty}(\mathcal{M})}(\Gamma^{\infty}(\mathcal{M},E))$, similiar to the isomorphism given in your question.
Plugging this in  yields
$$\Omega^{2}(\mathcal{M},\mathrm{End}(E))\cong  \mathrm{L}_{\mathrm{alt}}^{2}(\mathfrak{X}(\mathcal{M}),\mathrm{End}_{C^{\infty}(\mathcal{M})}(\Gamma^{\infty}(\mathcal{M},E)))$$
The module on the right hand side is naturally isomorphic to
the module of all  $C^{\infty}(\mathcal{M})$- multilinear maps from $\mathfrak{X}(\mathcal{M})\times\mathfrak{X}(\mathcal{M})\times\Gamma^{\infty}(\mathcal{M},E)$ to $\Gamma^{\infty}(\mathcal{M},E)$ which are antisymmetric in the first two slots and you have allready shown that $F_{\nabla}$ is such a map.
In summary $F\in\Omega^{2}(\mathcal{M},\mathrm{End}(E))$ is defined by $$F_p(X_p,Y_p)(s_p)=F_{\nabla}(X,Y,s)_p$$
where $X,Y,s$ are any sections of $TM,E$ which at $p$ equal $X_p,Y_p,s_p$.
