Let's assume we are in $\mathbb{R}^3$ and we have a $f(x,y) \in C^\infty$. Then a regular surface is given by $$M:=\{(x,y,z,)\in\mathbb{R}^3 | f(x,y)=z\},$$ and clearly we have a local parametrisation given by $$X(u,v) := (u,v,f(u,v)).$$
I now want to calculate the integral over the Gaussian curvature using the Gauss-Bonnet theorem.
My idea is that $$F(u,v,w):=X(u,v)+we_3$$ is a diffeomorphism between the $xy-plane$ and $M$ so the Euler characteristic of $M$ should be $$\chi(M)=2.$$
Using now the global Gauss-Bonnet theorem we get that for all $M$ induced by such an $f$ the integral over the Gaussian curvature satisfies $$\int_M KdA = 2\pi\chi(M)=4\pi.$$
But this seems strange to me because if we take for example the cone induced for $$f(x,y)=\sqrt{x^2+y^2},~~~and~~~z>0$$ $M$ should be flat and therefore the integral should vanish.
I am thankful for every helping hand.
There are at least two issues with your cone example:
$M$ is only a regular submanifold if zero is a regular value of the level set function $g(x,y,z) = f(x,y)-z$. This is not the case for the cone since $g$ is not differentiable at the cone point $(0,0,0)$. This is not a fatal issue, since one might imagine "rounding off" the cone's tip; however
Gauss-Bonnet holds only for compact surfaces, which the cone is not. You could restrict yourself to a graph over some compact region of the plane, in which case Gauss-Bonnet will hold, though you will need to include the geodesic curvature terms that arise at the boundary of the domain.