Contour integral of square root on its Riemann surface

Question

Consider a branch of the square root function $f(z)=z^{1/2}, z\in\mathbb{C}$ with $Im \thinspace {f(z)}>=0$, i.e.

                            $f(re^{i\phi}):=\sqrt{r}e^{i(\phi \thinspace mod \thinspace 2\pi)/2}.$

The function is not holomorphic at the origin and has a discontinuity on the non-negative real axis. Consequently, the contour integral of the function along the unit circle doesn't vanish. If we extend the function to it's Riemann surface then we'll get a single-valued continuous function, (informally $F(re^{i\phi})=\sqrt{r}e^{i\phi/2}$). The contour integral along the "double unit circle" $z(\phi)=e^{i\phi}, \phi\in{[0,4\pi]}$ on the Riemann surface now vanishes. I suspect this will be true for any other closed curve on the Riemann surface.

Is there a way to understand this in context of the residue theorem, suitably generalized to Riemann surfaces? Is it correct to say that the origin has now become an isolated (essential) singularity whose residue is zero and hence the integral of the function along any closed curve on the Riemann surface must always vanish?