Finding constant curvature metrics on surfaces for the case of positive Euler characteristic

If you like the case $K=-1$ better, one way to do this is to choose 3 points $\{x,y,z\} \subset S^2$, and use the uniformization theorem to find a complete conformally equivalent metric on $P=S^2-\{x,y,z\}$ with constant curvature $K=-1$. There is a unique such metric on $P$, which is conformally equivalent to $\mathbb{CP}^1-\{0,1,\infty\}$. Then fill in the punctures to get a conformally equivalent metric on $S^2$. One way to understand why uniformization of $S^2$ is a bit harder than the other cases is that there are Mobius transformations of $S^2$ which are conformal transformations but not isometries, so there is not a unique metric with $K=1$. By choosing three points, though, one gets rid of these conformal symmetries.

Addendum: Incidentally, the first proof that the Ricci flow on $S^2$ converges to the round metric was due to Bennett Chow, using a type of entropy defined specially on $S^2$ (Chow finished off a case not resolved by work of Hamilton). I think Perelman's work gives a new proof in the case of $S^2$, which I'm sure he realized, but I'm not sure has been properly disseminated. The idea is that if a singularity forms in finite time for Ricci flow on $S^2$, then one may take a rescaled limit to get a $\kappa$-non-collapsed positive curvature ancient solution (Perelman proof of this works in arbitrary dimensions, so is not special to $S^2$). In two dimensions, the only such solutions are solitons, which are either Hamilton's cigar or $S^2$ with the round metric (plus some non-orientable examples). But the cigar is collapsed, so the only possibility is $S^2$, which implies that the metric converges at the singular time to the round metric on $S^2$.

If you relax the fact the metrics $g$ and $g_0$ have to be point-wise conformal to be globally conformal, i.e there exists $\phi$ a diffeomorphism of $S^2$ such that $g=e^u \phi^*(g_0)$. Then the existence of a metric conform to $g_0$ with constant curvature is equivalente to the fact that there is only one conformal class on $S^2$. Which can be proved using the fact every surface is locally conformaly flat and the fact that $S^2$ is simply connected.