A question about Gauss Bonnet theorem If $S$ is a surface which is the complement of finitely many points in a compact surface, and the metric in $S$ is complete, then is Gauss-Bonnet theorem still valid for $S$?
No.

Take the sphere $S^2$ and remove one point, giving you a surface $S$ diffeomorphic to $\mathbb{R}^2$.  So we can equip $S$ with a metric making it isometric to $\mathbb{R}^2$, which is certainly complete.  $\mathbb{R}^2$ has zero curvature but its Euler characteristic is $1$, so
$$ 0 = \int_{\mathbb{R}^2} K dVol \ne \chi(\mathbb{R}^2) = 1.$$
No.

If you remove a finite set of points from a compact surface, it will no longer be complete. If you change the metric so that you do have a complete surface, you cannot apply the Gauss Bonnet theorem anyway, because the surface isn't compact.