This is just a question about what constitutes standard notation.
In a context where $A$ and $B$ are understood to be Hermitian matrices, what is usually meant by the bracket $[A,B]$ ?
On the one hand, I'd have thought it was standard for $[A,B]$ to denote the commutator $AB-BA$. On the other hand, I'd also have thought it was standard for $[-,-]$ to denote the bracket operation in a Lie algebra, in which case you'd want $[A,B]$ to be something like $i(AB-BA)$. These can't both be right. So is there some standard?
The answer to your question is, that for matrices $A$ and $B$ in $M_n(K)$ the standard Lie bracket is the commutator, i.e., $[A,B]=AB-BA$. The skew-hermitian matrices form a Lie subalgebra under this bracket, the unitary Lie algebra. For the commutator of Hermitian matrices see here. Then the usual commutator does not give a Lie subalgebra (the commutator of two Hermitian matrices is not Hermitian in general).
Every finite-dimensional Lie algebra can be faithfully represented by matrices with the standard commutator $[A,B]=AB-BA$, in some $\mathfrak{gl}(m)$ by Ado's theorem. However, $m$ might be large. In this sense, the usual commutator is standard. The given Lie algebra becomes a Lie subalgebra of $\mathfrak{gl}(m)$ with Lie bracket $[A,B]=AB-BA$.
Denote by $[A,B]$ always the Lie bracket of the Lie algebra, and the commutator of $A$ and $B$ always by $AB-BA$. Consider for example the Lie algebra $\mathfrak{su}(2)$, consisting of Hermitian matrices, with the Pauli matrices as basis. Here the standard commutator does not define a Lie bracket; a Lie bracket $[A,B]$ is rather given by $[A,B]:=i(AB-BA)$. However, $\mathfrak{su}(2)$ can also be represented by matrices inside $\mathfrak{gl}(3)$ by Ado's theorem, such that the Lie bracket is given by $[A,B]=AB-BA$. These matrices are given by the adjoint representation for $\mathfrak{su}(2)$ with the new basis $(e_1,e_2,e_3)$ and brackets $[e_1,e_2]=2e_3$, $[e_1,e_3]=-2e_2$, $[e_2,e_3]=2e_1$ - see also here. For example, $$ ad(e_1)=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -2 \\ 0 & 2 & 0\end{pmatrix}. $$