Angle created by a parallel transport along a smooth closed curve This answer uses the fact if you parallel transport a vector around a small smooth closed curve $C$, the angle through which it turns is $2\pi - \int_C \kappa_g(s)\,ds$. I don't see it immediately. I tried to use Gauss-Bonnet but it didn't work. A small proof will be very welcome.
 A: I assume you are working on a $2$-surface, and have chosen a coordinate on which the loops $C$ is going counterclockwise. Consider the unit tangent vector $T$ of $C$, and note when travel along $C$ once the vector $T$ turns by angle $2\pi$ (say, with respect to a nonzero, fixed, vector field on this coordinate system). Let $N$ be the outward pointing unit normal vector field, the geodesic curvature is defined by $k_g=-\langle \nabla_TT, N\rangle$, so that the Gauss Bonnet formula 
$$
\int_D K+\int_C k_g=2\pi
$$
holds; here $D$ is the disc bounded by $C$. 
Let $V$ be a vector at $C(0)$, and parallel translate $V$ along $C$, until you reach $C(L)=C(0)$, where 
$L$ is the length of $C$. Thus $\nabla_TV=0$. Let $s$ be the arclength parameter of $C$. Compute
$$
\frac{d}{ds}\langle V, T\rangle
=\langle \nabla_TV, T\rangle+\langle V, \nabla_TT\rangle=\langle V, \nabla_TT\rangle.
$$
Note $\nabla_TT\perp T$, so actually $\nabla_TT=-k_g N$. We can assume $|V|=1$. 
Let $\theta$ be the angle from $T$ to $V$, so 
$$
\frac{d}{ds}\langle V, T\rangle=\frac{d}{ds}\cos\theta
=\langle V, \nabla_TT\rangle=-k_g\langle V, N\rangle=-k_g\sin\theta. 
$$
Compute this out we see 
$\frac{d}{ds}\theta=-k_g.$
Integrate we get $\theta|_0^L=-\int_Ck_g$; recall $\theta=\angle T, V$ and $T$ rotated by $2\pi$, we see the rotation angle of $V$ is 
$$
2\pi+\theta\Big|_0^L=2\pi-\int_C k_g=\int_D K.
$$
