Maurer-Cartan form and Levi-Civita connection I am coming from this question, which has not being completely answered but I think is very interesting.
In several works ([Chern], [Griffiths] and [Clelland]) the Maurer-Cartan form for $E(n)$ is worked out in the following manner. They consider maps from $G=E(n)$ to $\mathbb{R}^n$, $x, e_1,\ldots, e_n$, and express their differentials in terms of the frame in which we are. But for me that doesn't seem natural because is something very particular of this example: the frame itself can be described in terms of the objects it describe. I consider more natural the general approach: the group $E(n)$ can be seen like a matrix group of a special type, that one with elements of the form
$$
\begin{pmatrix}
A & v\\
0 & 1\\
\end{pmatrix}
$$
with $A\in O(n)$ and $v\in \mathbb{R}^n$. And now you only have to apply the formula for MC form for a matrix group, $\theta=g^{-1}dg$, obtaining the same 1-forms.

Question 1 (solved)
Is this true for every Lie group of this type? That is, whenever we have a group $G\approx \mathbb{R}^n \rtimes H$ it can be seen as a subgroup of $GL(n+1)$ as above (see this QA in MSE) and we can interpret the columns as vectors in the homogeneous space $G/H\approx \mathbb{R}^n$. Then, does the Maurer-Cartan form tell us the variation of these vectors expressed in the current frame?

Back to the case of $E(2)$, for simplicity. The MC form is
$$
\theta=g^{-1}dg=\begin{pmatrix}
0&-d\theta&cos(\theta)da+sin(\theta)db\\
d\theta&0&-sin(\theta)da+cos(\theta)db\\
0&0&0&\\
\end{pmatrix}
$$
If we consider the basis of $\mathfrak{e}(2)$ given by
$$
B=\left\{
\begin{pmatrix}
0&0&1\\
0&0&0\\
0&0&0
\end{pmatrix},
\begin{pmatrix}
0&0&0&\\
0&0&1\\
0&0&0&\\
\end{pmatrix},
\begin{pmatrix}
0&-1&0&\\
1&0&0\\
0&0&0&\\
\end{pmatrix}
\right\}\equiv
$$
$$
\equiv\{\partial_a|_e,\partial_b|_e,\partial_{\theta}|_e\}
$$
we have
$$
\theta=\mu_1 \otimes\partial_a|_e+\mu_2 \otimes\partial_b|_e +\mu_3\otimes \partial{\theta}|_e
$$
with
$$
\mu_1=cos(\theta)da+sin(\theta)db
$$
$$
\mu_2=-sin(\theta)da+cos(\theta)db
$$
$$
\mu_3=d\theta
$$
In this case the Maurer-Cartan form has "two parts": $\mu_1, \mu_2$ on the one hand, and $\mu_3$ on the other hand. I think that $(\mu_1, \mu_2)$ corresponds to the canonical solder form and $\mu_3$ is the connection form of the Levi-Civita connection.

Question 2
Why is this the Levi-Civita connection? What relationship does it have (if any) with the group reduction of $GL(2)$ to $O(2)$ by means of the standard metric?

I have an intuition about some relation but I cannot grasp what it is... I know that the frame bundle for $\mathbb{R}^2$ is $\mathbb{R} \rtimes GL(2)$ and that the standard metric let us reduce the structure group of this principal bundle to $O(2)$...
References
[Chern]: Chapter 6 of S.S. Chern's book "Lectures on differential geometry"
[Griffiths]: On Cartan's method of Lie groups and moving frames as applied to uniqueness and existence questions in differential geometry
[Clelland]: From Frenet to Cartan: The Method of Moving Frames
 A: In question 1, I think you are just asking, if we have elements of $G$ written as $$g=\begin{pmatrix}h&v\\ 0&1\end{pmatrix}$$ then $$g^{-1}=\begin{pmatrix}h^{-1}&-h^{-1}v\\ 0&1\end{pmatrix}$$ hence $$g^{-1}dg=\begin{pmatrix}h^{-1}dh&-h^{-1}dv\\ 0&1\end{pmatrix}.$$ This is clearly since the Lie group operation is matrix multiplication.
Question 2: In books which discuss the moving frame, one learns that the soldering forms $\omega_1,\omega_2$ are precisely those for which the projection of orthonormal frame bundle to surface pulls back the dual of the orthonormal frame to $\omega_1,\omega_2$ and the Levi-Civita connection form $\omega_{12}$ is precisely the one for which $d\omega_1=-\omega_{12}\wedge\omega_2$ and $d\omega_2=\omega_{12}\wedge\omega_1$, so you can check that $\mu_3$ is the Levi-Civita.
For an example of such a book, my Introduction to Exterior Differential Systems discusses all of this in its many appendices, in particular in appendix G the structure equations are derived for Riemannian metrics on surfaces.
A: I have been working in question 2 and I think I have a good explanation. At the end is a triviality, but that's what (almost) always happens in math when you understand something.
I am going to discuss here the Euclidean plane from two different perspectives.
Euclidean plane
From the point of view of classical differential geometry
The Euclidean plane is the manifold $M=\mathbb{R}^2$, with coordinates $(x_1,x_2)$, together with the natural Riemannian metric
$$
g=dx_1\otimes dx_1+dx_2\otimes dx_2.
$$
It is, therefore, a Riemannian manifold.
Seen as a plain manifold we can consider that $M$ is endowed with a natural linear connection $\nabla$, such that $\nabla_{\partial_{x_i}}\partial_{x_j}=0\partial_{x_1}+0\partial_{x_2}$. The induced principal connection on the frame bundle is given by the 1-form
$$
\omega=
\begin{pmatrix}
c_{11}&c_{12}\\
c_{21}&c_{22}\\
\end{pmatrix}^{-1}\cdot\begin{pmatrix}
dc_{11}&dc_{12}\\
dc_{21}&dc_{22}\\
\end{pmatrix}\in \Omega(\mathbb R^2,\mathfrak{gl}(2))
$$
at
$$
f=
\begin{pmatrix}
c_{11} & c_{12}& x_1\\
c_{21} & c_{22}& x_2\\
0 & 0& 1\\
\end{pmatrix}\in FM
$$
Here it is shown how to construct $\omega$ from an arbitrary $\nabla$.
On the other hand, if we think of $M$ as a Riemannian manifold we can consider the Levi-Civita connection $\nabla_{LC}$. Since the metric is constant, the covariant derivative $\nabla_{LC}$ coincides with the natural covariant derivative $\nabla$, so it induces the same connection $\omega$ on $FM$.
But the metric $g$ also specifies a orthonormal frame bundle $OM$ (see here why). The elements of this principal bundle are
$$
f=
\begin{pmatrix}
c & -\sqrt{1-c^2}& x_1\\
\sqrt{1-c^2} & c& x_2\\
0 & 0& 1\\
\end{pmatrix}\in OM
$$
with $c\in [-1,1]$.
Since
$$
\omega|_{OM}=\begin{pmatrix}
c&\sqrt{1-c^2}\\
-\sqrt{1-c^2}&c\\
\end{pmatrix}\cdot\begin{pmatrix}
dc&\frac{c}{\sqrt{1-c^2}}dc\\
\frac{-c}{\sqrt{1-c^2}}dc&dc\\
\end{pmatrix}=
$$
$$
=\begin{pmatrix}
0&\frac{dc}{\sqrt{1-c^2}}\\
-\frac{dc}{\sqrt{1-c^2}}&0\\
\end{pmatrix} \in \Omega(\mathbb R^2,\mathfrak{o}(2)),
$$
according to Proposition 4.7. in vicenteBundles, this connection can be reduced to a connection on the orthonormal frame bundle determined by the metric $g$.
If we parameterize this principal bundle $OM$ with
$$
(x_1,x_2,\theta)\mapsto \begin{pmatrix}
\cos(\theta) & -\sin(\theta)& x_1\\
\sin(\theta) & \cos(\theta)& x_2\\
0 & 0& 1\\
\end{pmatrix}
$$
we obtain the more famous expression for $\omega$:
$$
\begin{pmatrix}
0&d\theta\\
-d\theta&0\\
\end{pmatrix}
$$
Remember: this 1-form tells us how much the bases at $f$ and $f'$ "fail to be constant" when we pass from the frame $f$ to a nearby frame $f'$, but expressing this mistake with respect to the frame $f$ itself.
From the point of view of Cartan geometry
The Euclidean plane is a Cartan geometry modeled over $(E(2),O(2))$ , indeed is the Klein geometry $(E(2),O(2))$. Moreover, it is a reductive Klein geometry since
$$
\mathfrak e(2)=\left\{\begin{pmatrix}
C & v\\
0 & 0\\
\end{pmatrix} 
:C\in \mathfrak{o}(2), v\in \mathbb R^2\right\}=\mathfrak o(2)\oplus \mathfrak p
$$
We have a natural choice for $\mathfrak p$
$$
\mathfrak{p}=\left\{\begin{pmatrix}
0 & p\\
0 & 0\\
\end{pmatrix} 
:p\in \mathbb R^2\right\}.
$$
With this in mind, remember that the Maurer-Cartan form describes all possible "infinitesimal displacements" of the frame $f$, but from the point of view of the frame $f$ itself. That is, if we pass from the frame $f$ to another frame $f'$, the Maurer-Cartan form at $f$ applied to the "vector" $\vec{ff'}=(dx_1,dx_2,d\theta)$ is a packet of information
$$
A=\begin{pmatrix}
0&-d\theta&cos(\theta)dx_1+sin(\theta)dx_2\\
d\theta&0&-sin(\theta)dx_1+cos(\theta)dx_2\\
0&0&0&\\
\end{pmatrix}\in \mathfrak e(2)
$$
Here is encoded, on the one hand, how much have we moved the base point of $f$ to the base point of $f'$ and, on the other, how much have we changed the basis itself. The natural choice of $\mathfrak p$ let us think that the information about the change of base point is in the $v$ part (the projection of the Maurer-Cartan form on $\mathfrak p$), and so the projection of the Maurer-Cartan form on $\mathfrak o(2)$, $\begin{pmatrix}0&d\theta\\-d\theta&0\\\end{pmatrix}$, tell us how much has the basis changed. That is, the same as the connection 1-form of the connection on the orthonormal bundle induced by the metric $g$ (which is the Levi-Civita connection).
To summarize:
In the orthonormal frame bundle induced by the metric $g$ we consider a displacement from a frame $f=\begin{pmatrix}C & p\\0 & 1\\\end{pmatrix}$ to a frame $f'=\begin{pmatrix}C' & p'\\0 & 1\\\end{pmatrix}$.
The principal connection $\omega$ induced by the Levi-Civita connection measures the change from $C$ to $C'$ as an element of $\mathfrak o(2)$.
The Cartan connection (Maurer-Cartan form) measures the change from $f$ to $f'$ as an element of $\mathfrak e(2)$. This change can be decomposed like the union of a change from $C$ to $C'$ and a change from $p$ to $p'$. This is reflected in the fact that  $\mathfrak e(2)=\mathfrak o(2)\oplus \mathfrak p$. If we focus on the change from $C$ to $C'$ we have the principal connection $\omega$.
