Actually, they are one and the same thing.
Before I delve into the question you asked, let me quickly describe a closely related analogy - the covariant derivative in GR. This is a quantity $\nabla_\mu$ that acts differently on different objects. In particular
$$
\nabla_\mu \phi = \partial_\mu \phi,~~~ (\nabla_\mu V)_\nu = \partial_\mu V_\nu - \Gamma^\lambda_{\mu\nu} V_\lambda,~~~ \cdots
$$
Here, $\phi$ is a scalar and $V$ is a vector, and so on.
Exactly as above, the covariant derivative on a gauge group acts differently depending on the representation of the field it acts on (incidently the same is true in GR, since $\phi$, $V$, etc. are classified as scalars and vectors under representations of $GL(n,{\mathbb R})$ which is the diffeomorphism group). For instance, if the field $F$ is a vector in representation $R_k$ of the gauge group $G$, then the covariant derivative acts as
$$
(D_\mu F)_i = \partial_\mu F_i - i A_\mu^a (T^a_k)_{ij} F_j ~~~~~~~...... (1)
$$
where $T^a_k$ is the generator of the Lie algebra ${\mathfrak g}$ corresponding to the representation $R_k$.
In particular, when one talks about a "matrix" in a representation one is really talking about a quantity in the adjoint representation. The adjoint representation is special because it can be described as a vector in the adjoint representation or as a matrix in any other representation $R_k$ as follows - A field in the adjoint representation has index $a$, i.e. $F^a$ (think gauge field). In this notation, it is a vector. However, it can analogously be described as a matrix as $F_{ij} = F^a (T^a_k)_{ij}$ in a representation $R_k$.
Now, let us apply our earlier definition of covariant derivative on the vector $F^a$ and see what it means for the matrix $F_{ij}$. First, since $F^a$ is a vector in the adjoint
$$
(D_\mu F)^a = \partial_\mu F^a - i A_\mu^b (T_{adj}^b)^{ac} F^c
$$
But, the generators in the adjoint representation are given by
$$
(T_{adj}^b)^{ac} = - i f^{bac}
$$
which implies
$$
(D_\mu F)^a = \partial_\mu F^a - A_\mu^b f^{bac} F^c = \partial_\mu F^a + A_\mu^b F^c f^{bca}
$$
Let us now go to the matrix notation by first contracting the above equation with $T^a_k$
$$
(D_\mu F)_{ij} = \partial_\mu F_{ij} + A_\mu^b F^c f^{bca} (T^a_k)_{ij}
$$
Now, we use the Lie algebra to write
$$
f^{bca} (T^a_k)_{ij} = - i [ T^b_k , T^c_k ] _{ij}
$$
which implies
$$
(D_\mu F)_{ij} = \partial_\mu F_{ij} - i A_\mu^b F^c [ T^b_k , T^c_k ] _{ij} = \partial_\mu F_{ij} - i [ A_\mu , F ]_{ij}
$$
or in simple matrix notation
$$
D_\mu F = \partial_\mu F - i [ A_\mu , F]
$$
Thus, we see here that the covariant derivative on a matrix in a representation $R_k$ is really the covariant derivative on a vector in the adjoint representation whose action is universally given by (1). They are therefore, one and the same thing.
