Suppose $v\colon \mathbb{R}^3\setminus\{x_1,\dots,x_n\}\to S^2$ is a smooth map, where all the points $x_i$ sit inside the unit ball $B^3$, and not on the boundary $S^2$. I suppose this map can be thought of as a unit vector field.
If we restrict $v|_{S^2}:S^2\to S^2$, we have a smooth map on continuous, compact, oriented $2$-manifolds, so $\deg(v|_{S^2})=\sum_{x\in v|_{S^2}^{-1}(q)}\operatorname{sgn}(x)$ where $q$ is a regular value, and $\operatorname{sgn}(x)=\pm1$ depending on whether $dF_x$ is orientation preserving/reversing.
What is the reason that this degree actually coincides with the sum of indices of $v$ at the deleted points $x_i$?
Because each index is itself the degree on a tiny sphere centered at the point. Moreover, if $f\colon W\to Y$ is a smooth map, $W$ a compact $(n+1)$-manifold with boundary, $Y$ a compact $n$-manifold, both oriented, then $\text{deg}(f|_{\partial W}\colon\partial W\to Y)=0$.