Notice that field strength transforms under adjoint representation. Moreover, recall that the connection piece of covariant derivative acts on an adjoint represented field in a commutator manner:
$D_\mu \phi = \partial_\mu \phi + [A_\mu, \phi] = \partial_\mu \phi^a T^a + f^{bca}A_\mu^b \phi^c T^a$.
With this understanding, the equivalence of the tensor notation and differential form notation looks to be immediate.
Reply to the comment:
Here is an argument why covariant derivative in adjoint representation should be in a commutator form.
1. By definition, a field $\phi$ transforms in adjoint representation of a Lie algebra in a form of
\begin{equation}
\delta \phi = [\phi, \lambda],
\end{equation}
where $lambda$ are infinitesimal Lie-algebra valued gauge parameters. The covariant derivative of a field in adjoint represention transforms again in adjoint representation, i.e. we will have
\begin{equation}
\delta D_\mu \phi = \left[ D_\mu\phi, \lambda \right],
\end{equation}
provided
\begin{equation}
D_\mu \phi := \partial_\mu \phi + [A_\mu, \phi].
\end{equation}
To show this, consider
\begin{align}
\delta D_\mu\phi
&= [\delta A_\mu , \phi ] + D_\mu \delta\phi \\
&= \left[ [A_\mu, \lambda], \phi \right] + [ \partial_\lambda , \phi ]
+ \partial_\mu [\phi, \lambda] + \left[ A_\mu, [\phi, \lambda]\right] \\
&= [\partial_\mu\phi, \lambda] + \left[ [A_\mu, \phi], \lambda \right] \\
&= [D_\mu\phi, \lambda],
\end{align}
where we used Jacobi identity as well as the transformation law of gauge field under gauge transformation,
\begin{equation}
\delta A_\mu = \partial_\mu \lambda + [A_\mu, \lambda].
\end{equation}
The gauge field strength transforms in adjoint representation because the action is then invariant:
\begin{equation}
\delta\, \text{tr}(F_{\mu\nu}F^{\mu\nu})
= 2 \text{tr}(F^{\mu\nu}[F_{\mu\nu}, \lambda]) = 0.
\end{equation}
Finally, notice that $\delta \phi = [\phi, \lambda]$ is equivalent to $(T_{a})_{bc} = f_{abc}$, because
\begin{equation}
\delta\phi = [ \phi^aT_a, \lambda^bT_b]
= \phi^a\lambda^b f_{abc} T^c
= \lambda^b(f_{bca})\phi^a T^c,
\end{equation}
then notice that this
\begin{equation}
\delta\phi^c = \lambda^b f_{bca}\phi^a
\end{equation}
is in a form of
\begin{equation}
\delta\phi^c = \lambda^b (T^{(R)}_b)_{ca} \phi^a,
\end{equation}
and thus the matrix elements of the generators are just structure constant in adjoint representation.
Answer to the question in the comment:
Let us assume
\begin{equation}
D_\mu \phi = \partial_\mu \phi + A_\mu\phi,
\end{equation}
as you suggested in the comment and original post. After taking the covariant derivative of the adjoint field $\phi = \phi^aT_a$, it should still in the adjoint representation and hence
\begin{equation}
(D_\mu\phi)_{be} = (D_\mu\phi)^a (T^{(\text{adj})}_a)_{be} = (D_\mu\phi)^a f_{abe}.
\end{equation}
Lets suppose the structure constant are totally antisymmetric (and neglect the subtlety for this to be true). Therefore, we see that the result of the covariant derivative should be antisymmetric in the matrix indices $b,e$. However, when we expand the covariant derivative, we get some trouble:
\begin{equation}
(D_\mu\phi)_{be} = \partial_\mu \phi^a f_{abe}
+ (A_\mu \phi)_{be}.
\end{equation}
While the first term above looks fine, the second term is problematic because
\begin{equation}
(A_\mu \phi)_{be}
= A_\mu^a \phi^c f_{acd}f_{bde} + (\phi A_\mu)_{be},
\end{equation}
where Jacobi identity is used. On the RHS, the first term is OK because it is antisymmetric under exchange of $b,e$, while the second term is making trouble.
However, you now see the way out. The simplest solution to get around is to move the second term to the LHS and so we get anti-symmetric $b,e$ on the RHS!
After we move the second term to the LHS, we get exactly the commutator and the covariant derivative
\begin{equation}
D_\mu\phi = \partial_\mu \phi + [A_\mu, \phi].
\end{equation}
