There seems to be a lot of confusion regarding the [,] operation. Well, the way I have learned it goes like this. Indeed, the two notations agree since the graded commutator [,] is defined as
$$
[α,β]=α\wedgeβ-(-1)^{pq}β\wedge α
$$
with $[α,β]=(-1)^{pq+1}[β,α]$ for $α\in \Omega^p(M,\mathfrak g)$ and $β\in \Omega^q(M,\mathfrak g)$, where $\mathfrak g$ is the Lie algebra of a Lie group $G$. Then, in your case
$$
DA=dA+\tfrac{1}{2}[A,A]=dA+\tfrac{1}{2}(A\wedge A-(-1)^{1\times 1}A\wedge A)=dA+A\wedge A
$$
Indeed, the use of $F=DA$ tends to be misleading sometimes due to the presence of 1/2. To that I agree, because in general one has in mind that $DB=dB+[A,B]$. Moreover, the Bianchi is quite short with this notation since
$$
DF=dF+[A,F]=\tfrac{1}{2}d[A,A]+[A,dA]+\tfrac{1}{2}[A,[A,A]]
$$
Well, $d[A,A]$ follows the usual derivation rule, i.e. $[dA,A]-[A,dA]=-2[A,dA]$ because $dA\in\Omega^2(M,\mathfrak g)$. Then, you can easily prove that $[A,[A,A]]=0$ (hint: the graded commutator satisfies a *graded Jacobi identity*). Taking the aforementioned properties into account, one directly sees that $DF=0$.

In an attempt to give some motivation for the introduction of the graded bracket, I think this has to do with a simple fact. Say that $α,β$ are just vector valued forms in $\omega^p(M,V)$ and $\Omega^q(M,W)$ respectively. Then, $$
α\wedge β=α^a\wedgeβ^be_a\otimes \tilde e_b
$$
where $e_a$ is a basis element of $V$ and $\tilde e_a$ a basis element of $W$. You see that the result lies in $\Omega^{p+q}(M,V\otimes W)$. Since the operation between Lie algebra elements is the Lie bracket, we can extend this to
$$
[α,\beta]=\alpha^a\wedge \beta^b[e_a,e_b]
$$
where for simplicity consider $e_a,e_b$ to be the generators of the algebra $\mathfrak g$ with $α,β$ as in the beginning (valued in this algebra). Since $[,]:\mathfrak g\times \mathfrak g\to \mathfrak g$, the result lies in $\Omega^{p+q}(M,\mathfrak g)$. The swapping rule is fairly straightforward since
$$
[α,\beta]=\alpha^a\wedge \beta^b[e_a,e_b]=(-1)^{pq}\beta^b\wedge\alpha^a[e_a,e_b]=-(-1)^{pq}\beta^b\wedge\alpha^a[e_b,e_a]=(-1)^{pq+1}[\beta,\alpha]
$$
Hope I helped a bit.

P.S: $A\wedge B$ is not the usual wedge product. If I remember correctly the clear notation is $A\wedge_{\rho}B$ where $(\rho,V)$ is a representation. Hence, say $A,B$ are $\mathfrak g$-valued. Then, we consider the adjoint representation, and we can write
$$
A\wedge_{\mathrm{ad}}B=A^a\wedge B^b\mathrm{ad}(e_a)e_b=A^a\wedge B^b[e_a,e_b]$$
This is why it makes sense to also have such operations between $\mathfrak g$-valued and $\mathfrak p$-valued forms if $\mathrm{ad}(\mathfrak g)\mathfrak p=[\mathfrak g,\mathfrak p]\subset \mathfrak g$ for example.

This post imported from StackExchange Physics at 2020-11-24 18:33 (UTC), posted by SE-user kospall