I've computed the Ricci tensor and the scalar curvature of the following metric on $I\times S^2, I\subset \mathbb{R}$:
$$g=(1-r^2)^{-1}dr \otimes dr + r^2 d\theta \otimes d\theta + r^2\sin^2\theta d\phi \otimes d\phi$$
My orthonormal (co)frame is $\omega^r=(1-r^2)^{-\frac{1}{2}}dr$, $\omega^\theta=rd\theta$ and $\omega^\phi=r\sin\theta$ $d\phi$. Using the three Cartan structure equations I managed to calculate the connection forms and the curvature forms. From the curvature forms I calculated the non-vanishing components of the curvature tensor $R_{ijk}^l$ on this frame. Finally I used $R_{ij}=R_{kij}^k$ (Einstein summation convention implied) to obtain the non-vanishing components of the Ricci tensor, in this case they are $R_{rr}$, $R_{\theta\theta}$ and $R_{\phi\phi}$, and then $S=g^{ij}R_{ij}$ was used for the scalar curvature, where the Einstein summation convention is once again implied.
In order to check my calculations I went and made a short notebook in Mathematica that computes the Ricci tensor and scalar for a given metric the traditional way (i.e. by computing the Christoffel symbols of the Levi-Civita connection), and the results don't agree. It turns out that the components of the Ricci tensor I'm getting by following Cartan's method are already multiplied by the inverse metric, i.e. $R_{rr}$, $R_{\theta\theta}$ and $R_{\phi\phi}$ are actually the components of the mixed Ricci tensor and not the covariant Ricci tensor. I just have then to sum them to obtain the right expression for the scalar curvature $S$, without further multiplications by the inverse metric.
This is what I don't get: I thought I was calculating the components of the covariant Ricci tensor, but, in fact, as the notebook showed me, what Cartan's method gave me was the components of the mixed tensor. Why is that so? Why does the inverse metric come already factored in the result?