Moving frames method for non-matrix Lie group I am having troubles in understanding the modern definition of moving frames method. 
Classically, the idea of moving frames is "to express the variation in terms of the moving frame itiself". This is clear if we have a matrix Lie group $G$, like $O(n), GL(n)$...
But if $G$ is not a matrix group, we cannot express the variation in terms of the moving frame. At this point are introduced Lie algebras, Maurer-Cartan forms and all things that I understand but I cannot figure out where they come from. 
If $G$ is a matrix Lie group, the Maurer-Cartan form is the usual that I know and so is the Maurer-Cartan equation and that's ok. My question is: why is this the correct generalization of moving frame method to "abstract" Lie group? What's the intuition behind this? Who does this generalization?
EDIT
In particular I am asking what's the intuition behind these generalizations:


*

*From the matrix Maurer-Cartan form $\omega = g^{-1}dg$ to the "the unique left-invariant $\mathfrak{g}$-valued 1-form on $G$ such that $\omega|_e : T_eG \to \mathfrak{g}$ is the identity.

*From the matrix Maurer-Cartan equation $\text{d}\omega = -\omega\wedge\omega$ to the equation $\text{d}\omega = -\frac{1}{2}[\omega, \omega]$.

*Why when need the Lie bracket if before we don't?


EDIT II
I tought this answers for 1. In "classic" moving frames method, $G$ acts on every $T_gG \cong \mathbb{R}^{n, n}$ because $G$ is a matrix group. But this is true also for an abstract Lie group $G$. In fact we have the left action of $G$ on $TG$ given by
\begin{align}
\cdot: G \times TG &\to TG \\
(g, v) &\mapsto \text{d}L_g(v)
\end{align}
So if we have a moving frame $f: M \to G$ we also have that $df = f \cdot \omega_f$ with $\omega_f: TM \to TG$. It's easy to see that really we have that $\omega_f: TM \to T_eG$ and also that $\omega_f = f^*\omega$ with $\omega: TG \to T_eG$. This is the Maurer-Cartan form, but this approach is for me the same as Cartan's; for me defining the Maurer-Cartan form as "the unique..." breaks the intuition behind the method. 
In these terms I would reformulate question 2. Classically, we start from $dg = g\omega$ and we exterior-differentiate to get $0 = \text{d}g\wedge\omega + g\text{d}\omega = g(\omega \wedge \omega + \text{d}\omega)$ and so the Maurer-Cartan equation. My question is: can we reply this approach starting from $df = f \cdot \omega_f$? For me makes sense that we have to use Lie bracket since is the analogous of $\text{d}^2$ in the antisymmetrization process: but how?
 A: I am not clear about what you find confusing. The answer to the first question seems simply to be that the expression $g^{-1}dg$ is meaningless on an abstract Lie group. To be more precise, the expression $g$ is subjected to the exterior derivative $d$ by first being interpreted as a function on the group $g \colon G \to GL(n)$, i.e. the inclusion function, if $G\subseteq GL(n)$. But on an abstract Lie group, there is no such inclusion function. So we need to find a property of $g^{-1}dg$ that holds on matrix Lie groups $G\subseteq GL(n)$, and which can generalize to all Lie groups.
I think Deane answered the other questions. For more detail, the definition of a wedge product of 1-forms $\alpha,\beta$ valued in an arbitrary algebra $A$ (not necessarily a Lie algebra) with multiplication $x\in A, y \in A \mapsto m(x,y)\in A$ is $m(\alpha,\beta)(v,w)=m(\alpha(v),\beta(w))-m(\alpha(w),\beta(v))$. I don't think there is any more reasonable definition. When applied to the Lie algebra of matrices under bracket, this yields $[\alpha,\beta]=2\alpha\wedge\beta$, if you expand it out. Hence the factor of 2 to relate Lie bracket to matrix multiplication.
We need the Lie bracket because, as Deane says, there is no other obvious product in the Lie algebra.
Edit: maybe I didn't give enough intuition. The intuition behind left translation is clear. Left translation identifies all tangent spaces of any Lie group. But that makes the definition of Maurer-Cartan form obvious: you give Maurer and Cartan your velocity as a vector in a tangent space of $G$, and they hand you back the corresponding velocity vector at the identity element, so you can compare velocities at different points. But that is just the definition of Maurer-Cartan form, and is clearly just $g^{-1} dg$ on a matrix Lie group: the $g^{-1}$ in $g^{-1}dg$ is the left translation back to the identity.
We derive the equation for $d\omega$ as in lots of books: $d\omega(X,Y)=L_X(\omega(Y))-L_Y(\omega(X))-\omega([X,Y])$, applied to left invariant vector fields, using the left invariance of $\omega$ to kill the first two terms. 
I insist that the superior approach to the study of moving frames is to avoid ever taking a local section $f$. Instead, to each immersed submanifold $\iota \colon M \to X=G/H$, we associate the pullback bundle $\iota^*G$, and the pullback Maurer-Cartan form. We think of this pullback bundle as the bundle of adapted frames (or coframes). This generalizes easily to immersed submanifolds in Cartan geometries and to foliations of Cartan geometries. Whatever you are reading about the moving frame method, throw your source away and work it out from the beginning using bundles and reduction of bundles.
A: Here is an abstract formulation of the equation $df = f\omega$ on the frame bundle, which is assumed to be the Lie group $G$.
Let $g: G \rightarrow G$ be the identity map. Then $dg: T_*G \rightarrow T_*G$ is also the identity map. For each $g$, we can compose this with the differential of
left translation $dL_{g^{-1}}(g): T_gG \rightarrow T_eG = \mathfrak{g}$. This defines the left invariant Maurer-Cartan form as a $\mathfrak{g}$-valued $1$-form on $G$,
$$
  \omega = dL_{g^{-1}}(g)\circ dg: T_gG \rightarrow \mathfrak{g}.
$$
Equivalently,
$$
  dg = dL_g(e) \circ \omega: T_gG \rightarrow T_g.
$$
This is formally equivalent to
$$
dg = g\omega.
$$
I have not worked out the details of how to derive the Maurer-Cartan equation from the last equation, but note that to differentiate $dg$, you need a connection on the tangent bundle of $G$. However, the Maurer-Cartan equation does not use any connection. I believe it is derived using a left invariant torsion-free connection, and then you verify that it in fact does not depend on the connection used.
