Why the well-defined of Gauss map depends on surface is orientable?

Question

Let $S$ is a surface. Define a mapping $g:S\rightarrow S^2\subset R^3$ of $S$ into the unit sphere $S^2$ , associating to every $p\in S$ a unit vector $N(p)\in S^2$ normal to $T_pS$. Why the well-defined of $g$ depends on $S$ is orientable ?

Besides, if $S$ is a n-dim manifold ,then the range of $g(S)$ is $S^n$ ?

Answer 1

If the surface is not orientable, then the unit normal vectors $N(p)$ are only defined up to sign. There is no way to consistently decide whether to take one normal vector or its negative.