Curvature Matrix on a Smooth Vector Bundle Suppose $\pi:E\rightarrow M$ is a smooth vector bundle with a metric connection $\nabla$, and let $(\sigma_i)$ be a local frame over some open $U\in M$.
Why is the curvature
\begin{equation}
R(X,Y)\sigma_i=\nabla_X\nabla_Y\sigma_i-\nabla_Y\nabla_X\sigma_i-\nabla_{[X,Y]}\sigma_i
\end{equation}
a $\text{Hom}(E,E)$-valued 2-form? Further, how does this translate into a matrix form
\begin{equation}
\Omega(\sigma_i)=\Omega_{ij}\sigma_j,
\end{equation}
where $\Omega_{ij}$ is an antisymmetric matrix? I had thought I understood how to construct $\Omega_{ij}$, but I can't come up with any reason for
\begin{equation}
\Omega_{ij}=-\Omega_{ji}
\end{equation}
to be true, unless there is some way of indicial symmetry which I don't see for the Christoffel symbols, defined here by
\begin{equation}
\nabla_X\sigma_i=\Gamma_{ik}^jX^k\sigma_j,
\end{equation}
where $X$ is a smooth vector field on $M$.
 A: 
Why is the curvature 
  \begin{equation}
R(X,Y)\sigma_i=\nabla_X\nabla_Y\sigma_i-\nabla_Y\nabla_X\sigma_i-\nabla_{[X,Y]}\sigma_i
\end{equation} 
  a $\text{Hom}(E,E)$-valued 2-form?

The point is that the map $(X,Y,\sigma) \mapsto R(X,Y)\sigma$ is multilinear over $C^\infty(M)$, which you can check by inserting an expression like $f_1 X_1 + f_2 X_2$ in place of $X$ and checking that all the terms involving derivatives of $f_1,f_2$ cancel out. Because it depends antisymmetrically on $X,Y$, and the result for any fixed $X,Y$ is a map from $\Gamma(E)$ to itself that's linear over $C^\infty(M)$, this shows that it's actually a smooth $\text{End}(E)$-valued $2$-form. The basic ideas are spelled out in the proof of Lemma 10.29 in my Introduction to Smooth Manifolds, 2nd edition.

Further, how does this translate into a matrix form
  \begin{equation}
\Omega(\sigma_i)=\Omega_{ij}\sigma_j,
\end{equation}
  where $\Omega_{ij}$ is an antisymmetric matrix? 

The formula you wrote is really a shorthand for
$$
R(X,Y)\sigma_i = \Omega_{ij}(X,Y)\sigma_j.
$$
This defines $\Omega_{ij}$ for each $i$ and $j$ as a $2$-form on $U$. It will be antisymmetric in $i,j$ provided the frame $\boldsymbol{\sigma_j}$ is orthonormal. I don't have time to write out a complete proof now, but it follows from the fact that with respect to an orthonormal basis, the connection $1$-forms take values in the Lie algebra $\mathfrak{so}(n)$. This in turn follows from the fact that the connection is metric-compatible.  
