Consider a gauge theory with gauge group $GL(n,R)$.

First of all, let me remind you the basics of gauge transformations:

Let $G$ be a gauge group, $g(x)∈ G$ be an element of $G$. Then:

$\psi (x)→g(x)\psi (x)$

$A_\alpha→g(x)A_\alpha g^{-1}(x)-\frac{∂g(x)}{∂x^{\alpha}}g^{-1}(x)$

is a gauge transformation, and covariant derivative defined as 

${\nabla}_{\alpha}\psi = \frac{∂\psi}{∂x^{\alpha}}+A_\alpha \psi$

Now consider coordinates $(x^1, ..., x^n)$ in region $U$. They define basis of vectors space $\frac{∂}{∂x^1},...,\frac{∂}{∂x^n}$. So, tangent vector fields in region $U$ can be considered as vector-valued functions: $\xi = (\xi^1,...,\xi^n)$.
Change of coordinates in $U$: $x^{\nu}→x^{{\nu^{\prime}}}=x^{{\nu^{\prime}}}(x)$ defines local transformation:

$\xi^\nu→\xi^{\nu^\prime} = \frac{∂x^{\nu^\prime}}{∂x^\nu}\xi^\nu = g(x)\xi$.

Here matrix $g(x) = (\frac{∂x^{\nu^\prime}}{∂x^\nu})$ belongs to $GL(n,R)$, and inverse matrix has the form $g^{-1}(x)=(\frac{∂x^\nu}{∂x^{\nu^\prime}})$.

Lie algebra of $GL(n,R)$ is formed by all matrices of degree $n$, so the "gauge field" $A_\mu (x)$ is also matrix of degree $n$. Let us denote it's elements as follows:

$(A_\mu)^{\nu}_{\lambda}=\Gamma^{\nu}_{\lambda \mu}$.

Covariant derivative of the vector $\xi$ reads as follows:

$(\nabla_{\mu}\xi)^\nu=\frac{∂\xi^\nu}{∂x^\mu}+\Gamma^{\nu}_{\lambda \mu}\xi^\lambda ↔ \nabla_\mu \xi=\frac{∂\xi}{∂x^\mu}+A_\mu \xi$ (right side is in matrix form!)

There is only one thing left to check, namely the form of gauge field transformation.

Using general rule of transforming gauge field we obtain:

$\Gamma^{\nu}_{\lambda \mu}→\Gamma^{\nu^\prime}_{\lambda^\prime \mu}=\frac{∂x^{\nu^\prime}}{∂x^\nu}\Gamma^{\nu}_{\lambda \mu}\frac{∂x^\lambda}{∂x^{\lambda^\prime}}+\frac{∂x^{\nu^\prime}}{∂x^\nu}\frac{∂}{∂x^\mu}(\frac{∂x^\nu}{∂x^{\lambda^\prime}})$.

Since $A_\mu$ is a covariant vector, then $A_{\mu^\prime}=\frac{∂x^\mu}{∂x^{\mu^\prime}}A_\mu$. Hence we obtain:

$\Gamma^{\nu^\prime}_{\lambda^\prime \mu^\prime}=\frac{∂x^\mu}{∂x^{\mu^\prime}}\frac{∂x^{\nu^\prime}}{∂x^\nu}\Gamma^{\nu}_{\lambda \mu}\frac{∂x^\lambda}{∂x^{\lambda^\prime}}+\frac{∂x^{\nu^\prime}}{∂x^\nu}\frac{∂^2 x^\nu}{∂x^{\lambda^\prime}∂x^{\mu^\prime}}$.

Q.E.D.

And final remark: the commutator of two covariant derivatives leads to expression of the Riemann tensor:

$(F_{\mu\nu})^\rho_\lambda = R^\rho_{\lambda ,\mu\nu}$