Just to test out the Cartan formalism, I decided to apply it to the sphere. So, it admits a metric,
$$\mathrm{d}s^2 = \mathrm{d}r^2 + r^2 \sin^2 \phi \mathrm{d}\theta^2 + r^2 \mathrm{d}\phi^2$$
from which I may define an orthonormal basis,
$$\omega^{r} = \mathrm{d}r, \, \, \, \ \omega^{\theta} = r\sin \phi \mathrm{d}\theta, \, \, \, \, \omega^{\phi} = r \mathrm{d}\phi.$$
I applied Cartan's first equation, i.e.
$$\mathrm{d}\omega^a + \Gamma^a_b \wedge \omega^b = 0$$
to determine the connections $\Gamma^{a}_b$; I obtained
$$\Gamma^{\theta}_\theta = -\frac{1}{r}\omega^r - \frac{1}{r} \cot \phi \, \omega^{\phi} \, \, \, \, , \, \, \, \, \Gamma^{\phi}_\phi = -\frac{1}{r} \omega^r$$
in terms of the orthonormal basis. Cartan's second equation states,
$$R^a_b = \mathrm{d}\Gamma^a_b + \Gamma^a_c \wedge \Gamma^c_b$$
but all exterior derivatives of the connections vanish. In addition, I think the second term vanishes as well, implying the curvature 2-form $R^a_b = 0$. However, this is not the correct result? By the Chern-Gauss-Bonnet theorem, this would imply the Euler characteristic $\chi = 0,$ which implies the genus $g=1$. But a sphere has genus $g=0$. Can someone see where I went wrong in the calculation?
What you did is not on a sphere. You performed all the calculations in $\mathbb{R}^3$, which is flat, only in spherical coordinates!
Try instead dropping the $dr$, since you don't need it. Stay on the sphere. You get a metric: $$ ds^2 = r^2 sin^2\theta\, d\phi^2 + r^2\,d\theta^2. $$
With only two dimensions, calculations get much, much simpler, try! (For example, 2-forms are 1-dimensional...)
To simplify calculations even more, you can even fix $r=1$ (don't do this is if you want to find the relation between $r$ and the curvature).