Let me see if I remember correctly. To be the Levi Civita connection, the connection forms must satisfy the following equations:
TorsionLess
$$\mbox{d}\theta^{i}={\sum}\theta^{j}\wedge\theta_{j}^{i}$$
Compatibility with the metric
$$\mbox{d}g_{ij}={\sum}\left(\theta_{i}^{k}g_{kj}+\theta_{j}^{k}g_{ki}\right).
$$
I think Cartan structure equations gives you the first condition and not the second.
I'll explain everything with a concrete example you can then generalize easily. Let us consider the Heisenberg group. We have that the Maurer Cartan form is given by
$$A^{-1}\mbox{d}A=\left(\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right)\mbox{d}x+\left(\begin{array}{ccc}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right)\left(\mbox{d}y-x\mbox{d}z\right)+\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}\right)\mbox{d}z,$$ so we have that the invariant forms are
$$\theta^{1}=\mbox{d}x, \theta^{2}=\mbox{d}y-x\mbox{d}z, \theta^{3}=\mbox{d}z. $$
If we apply the external derivative we obtain
$$\mbox{d}\theta^{1} =0,$$
$$\mbox{d}\theta^{2} =-\mbox{d}x\wedge\mbox{d}z=-\theta^{1}\wedge\theta^{2},$$
$$\mbox{d}\theta^{3} =0.$$
The dual vectors of theinvariant differential form $\theta^{1}$, $\theta^{2}$ and $\theta^{3}$ are the vectors
$$\tilde{E}_{1} = \frac{\partial}{\partial x},$$
$$\tilde{E}_{2} = \frac{\partial}{\partial y},$$
$$\tilde{E}_{3} = x\frac{\partial}{\partial y}+\frac{\partial}{\partial z}. $$
So in the moving frame $ \left\{ \tilde{E}_{1},\,\tilde{E}_{2},\,\tilde{E}_{3}\right\} $ we have that $\left(g_{A}\right)_{ij}$ is the identity and then the equations for the compatibility with the metric (if we want to find the Levi Civita Connection) are
$$\omega_{j}^{i}+\omega_{i}^{j}=0.$$
Then from the structure equations we have
$$\mbox{d}\theta^{1}= 0 =\left(\mbox{d}y -x\mbox{d}z\right)\wedge\omega_{2}^{1}+\mbox{d}z\wedge\omega_{3}^{1},$$
$$\mbox{d}\theta^{2}= -\mbox{d}x\wedge\mbox{d}z =-\mbox{d}x\wedge\omega_{2}^{1}+\mbox{d}z\wedge\omega_{3}^{2},$$
$$\mbox{d}\theta^{3}= 0 =-\mbox{d}x\wedge\omega_{3}^{1}-\left(\mbox{d}y-x\mbox{d}z\right)\wedge\omega_{3}^{2}.$$
If we solve we then find the connection forms
$$\omega_{2}^{1} =-\omega_{1}^{2} =\frac{1}{2}\mbox{d}z=\frac{1}{2}\theta^{3},$$
$$\omega_{3}^{2} =-\omega_{2}^{3} =\frac{1}{2}\mbox{d}x=\frac{1}{2}\theta^{1},$$
$$\omega_{3}^{1} =-\omega_{1}^{3} =\frac{1}{2}\left(\mbox{d}y-x\mbox{d}z\right)=\frac{1}{2}\theta^{2}.$$
And then you have the curvature forms
$$\Omega_{2}^{1}= \theta_{3}^{1}\wedge\theta_{2}^{3} =\frac{1}{4}\theta^{1}\wedge\theta^{2},$$
$$\Omega_{3}^{2}= \theta_{1}^{2}\wedge\theta_{3}^{1} =\frac{1}{4}\theta^{2}\wedge\theta^{3},$$
$$\Omega_{3}^{1}= -\frac{1}{2}\theta^{1}\wedge\theta^{3}+\theta_{2}^{1}\wedge\theta_{3}^{2} =-\frac{3}{4}\theta^{2}\wedge\theta^{3}.$$
And at the end the Riemann coefficients
$$R_{212}^{1} = \frac{1}{4},$$
$$R_{323}^{2} = \frac{1}{4},$$
$$R_{323}^{1} = -\frac{3}{4}.$$
I have computed that the 1-form $\omega:= p^*\omega' \in \Omega^1(Q\times H', \mathfrak{h'})$ descends to a form on $Q\times_{\alpha}H'$
I don't think this is true: firstly as Ted points out, the Lie group homomorphism $\alpha:H\to H'$ induces a Lie algebra homomorphism $\dot{\alpha} = T_e\alpha:\mathfrak{h}\to\mathfrak{h}'$. For $(q,h')\in Q\times H'$ and $\xi\in\mathfrak{h}$, the vector along the quotient direction
$$
(q\cdot \xi, -\dot\alpha(\xi)\cdot h')\in T_{(q,h')}(Q\times H')
$$
when contracted with $\omega = p^*\omega'$ takes the value $-\dot\alpha(\xi)$, which is non-zero for $\xi\notin \ker \dot\alpha$. However this vector descends to the zero vector on $Q\times_\alpha H'$.
I haven't worked through the details, but I think the correct form to consider would be
$$
\omega = \dot\alpha(p_1^*\omega_Q) + p_2^*\omega',
$$
where $p_1:Q\times H'\to Q$ and $p_2:Q\times H'\to H'$ denote the respective projections. It doesn't suffer from the above problem: applying it to the quotient direction $(q\cdot \xi, -\dot\alpha(\xi)\cdot h')$ yields 0, and so it should descend to a 1-form on $Q\times_\alpha H'$, which will be a connection.
Edit: I went back to work out the details of this, and it seems I was incorrect in my guess. The correct form should be
$$
\omega_{(q,h')} :=\mathrm{Ad}_{(h')^{-1}}\left(\dot{\alpha}(p_1^*\omega_Q )_{(q,h')}\right) + (p_2^*\omega_L')_{(q,h')}=\mathrm{Ad}_{(h')^{-1}}\left(\dot{\alpha}(p_1^*\omega_Q )_{(q,h')} + (p_2^*\omega_R')_{(q,h')}\right)
$$
Here $\omega_L' (\omega_R')$ is the left-(right-)invariant Maurer-Cartan form on $H'$. Letting $\pi_\alpha:Q\times H' \to Q\times_\alpha H'$ denote the quotient map,
$\omega$ vanishes on vectors of the form $(q\cdot\xi, -\dot\alpha(\xi)\cdot h')$ (i.e. vectors along the fibres of $\pi_\alpha$), and is invariant under the quotient action. So it reduces to a form $\omega_\alpha$ on $Q\times_\alpha H'$
$$
\omega = \pi_\alpha^*\omega_\alpha.
$$
The obvious right actions $R_\cdot$ and $\check{R}_\cdot$ of $H'$ on $Q\times H'$ and $Q\times_\alpha H'$ commute with $\pi_\alpha$,
$$
\pi_\alpha\circ R_{h'} = \check{R}_{h'}\circ \pi_\alpha,
$$
and it is straightforward to verify that $R_{k'}^*\omega = \mathrm{Ad}_{(k')^{-1}}\omega$ for all $k'\in H'$. It follows that
$$
R_{k'}^*\pi_\alpha^*\omega_\alpha = \mathrm{Ad}_{(k')^{-1}}\pi_\alpha^*\omega_\alpha \implies \pi_\alpha^*\check{R^*}_{k'}\omega_\alpha = \pi_\alpha^*(\mathrm{Ad}_{(k')^{-1}}\omega_\alpha).
$$
Since $\pi_\alpha$ is a submersion, this implies $\check{R^*}_{k'}\omega_\alpha = \mathrm{Ad}_{(k')^{-1}}\omega_\alpha$. Also $\omega_{(q,h)}((q,h)\cdot \xi) =\omega_{(q,h)}(0_q,h\cdot \xi) = \xi$ and the fact that $\pi_\alpha$ commutes with the right action implies the equivalent result for $\omega_\alpha$. So $\omega_\alpha$ is a connection form on $Q\times_\alpha H'$.
Best Answer
Let $h$ be any point of $G$ and let $Y_h\in T_hG.$ Then if $g\in G$ we have \begin{equation} R_{g}^{*}(\omega)(Y_h)=\omega((R_{g})_{*}Y_h)=(L_{g^{-1}h^{-1}})_{*}((R_{g})_{*}Y_h)\overset{(*)}{=}(R_{g})_{*}(L_{g^{-1}h^{-1}})_{*}(Y_{h}) \end{equation} $$=(R_{g})_{*}(L_{g^{-1}})_{*}((L_{h^{-1}})_{*}Y_{h})=(R_{g})_{*}(L_{g^{-1}})_{*}(\omega(Y_h))=Ad(g^{-1})(\omega(Y_{h})). $$ The equality $(*)$ follows from the commutation of $R_{a}$ and $L_{b}$, for all $a,b\in G.$