Gauge covariant derivative on form Let $e$ be a one-form gauge field that belongs to the adjoint representation of the gauge group, that is SO(1,2). It is defined as 
\begin{equation}
e = e_{\alpha}^{A}T_Adx^{\alpha}.
\end{equation}
The $T_A$ are the generators of the Lie algebra SO(1,2) and obey the usual commutation relations 
\begin{equation}
[T_A,T_B] = -\epsilon_{ABC}T^C,
\end{equation}
with $A,B,C=0,1,2$ and $\epsilon_{012}=1$. The group indices $A,B,C$ are raised and lowered with the flat metric $ \eta_{A,B}=diag(1,-1,-1)$.
The covariant derivative is defined as \begin{equation}
D = d+[e, \quad]\end{equation}
The field strengh is defined in terms of the commutator and it yields 
\begin{equation}
[D_{\alpha},D_{\beta}] = F_{\alpha \beta}^{A}T_{A}
\end{equation}
It is explicity given by 
\begin{equation}
F_{\alpha \beta} = \partial_{\alpha}e_{\beta}^{A}-\partial_{\beta}e_{\alpha}^{A}-\epsilon_{BC}^{A}e_{\alpha}^Be_{\beta}^C
\end{equation}
Question
I am used to the usual notation in term of coordinates but I am lost here. What shall I put in the commutator ? A random 1-form ? How to explicitely get the last result by evaluating the commutator ? 
 A: Using the Lie algebra valued function:
$$e_{\alpha} = e_{\alpha}^AT_A$$
We can write the covariant derivative components
$$D_{\alpha} = \partial_{\alpha} + \mathrm{ad}(e_{\alpha})$$
where, $ \mathrm{ad}(X) = [X, .]$ is the adjoint representation. Please notice that it is linear in the components of $e_{\alpha}$
Thus
$$ \begin{align*}
[D_{\alpha}, D_{\beta}] &= \mathrm{ad}(e_{\alpha}) \partial_{\beta}- \partial_{\beta} \mathrm{ad}(e_{\alpha}) - \mathrm{ad}(e_{\beta}) \partial_{\alpha}+ \partial_{\alpha} \mathrm{ad}(e_{\beta}) +  [\mathrm{ad}(e_{\alpha}) , \mathrm{ad}(e_{\beta}) \\
&= -\mathrm{ad}(\partial_{\beta}e_{\alpha}) + \mathrm{ad}(\partial_{\alpha}e_{\beta}) + \mathrm{ad}([e_{\alpha} , e_{\beta}]) \\
&= \mathrm{ad}(\partial_{\alpha}e_{\beta} -\partial_{\beta}e_{\alpha} + [e_{\alpha} , e_{\beta}]) 
\end{align*}
$$
where the linearity of the adjoint representation of the Lie algebra was used in the application of Leibniz rules and the fact that it is a representation: $[\mathrm{ad}(X), \mathrm{ad}(Y)] = \mathrm{ad} ([X,Y])$  
The result is evident from the last expression.
A: 
The covariant derivative is defined as 
  $$
\begin{equation}
D = d+[e, \quad]\end{equation}
$$
  What shall I put in the commutator ? A random 1-form ?

For the sake of generosity, let's look at an n-form $X$ which transforms covariantly under some gauge transformation $g$
$$
X \rightarrow gXg^{-1}.
$$
The covariant derivative of $X$ should be
\begin{equation}
DX = dX+AX-(-1)^n XA,
\end{equation}
where $A$ is the gauge field 1-form corresponding to the gauge transform $g$ and wedge $\wedge$ product between forms is assumed. The first $-$ sign in the above expression comes from $g^{-1}$ in $gXg^{-1}$. The second $(-1)^n$ sign has to do with the wedge $\wedge$ product (anti)commuting properties between 1-form $d$ and n-form $X$.
So for even(0, 2, ...)-form $X$, the covariant derivative is
$$
\begin{equation}
DX = dX+[A, X] = dX+AX- XA.
\end{equation}
$$
And for odd(1, 3, ...)-form $X$, the covariant derivative is
$$
\begin{equation}
DX = dX+[A, X]_+ = dX+AX+ XA.
\end{equation}
$$
For example, the covariant derivative of the tetrad/vielbein/vierbein 1-form $e$ is 
$$
T= D_\omega e = de+[w, e]_+  = de + \omega e + e\omega,
$$
where $\omega $ is the spin connection 1-form, which is the gauge field corresponding to the local Lorentz gauge transformation $g_{Lorentz}$. Incidentally, $T$ is the torsion 2-form. 
