Consider the following lemma
(1) How come he talks about degrees here, after all he doesn't assume $X$ to be oriented? (2) Why is $\bar{v}|\partial X$ homotopic to $g$? (NOTE: we consider them as maps to $S^{m-1}$ isn't it? This is very important)
I have thought about these questions on my own, but I don't want to give my own answers as I want yours to be independent of my mine (my answers are plausible, so they might convince you even if they are actually wrong).
For question (1), the first sentence that you quote says "Let $X \subset \mathbb{R}^m$ be a compact $m$-manifold with boundary". Since $X$ has the same dimension as $\mathbb{R}^m$, $X$ naturally inherits an orientation from $\mathbb{R}^m$: at a point $p \in X$ we have equality of tangent spaces $T_p X = T_p \mathbb{R}^m$, and so we just use the natural orientaton on $\mathbb{R}^m$ itself.
For question (2), $v$ is by assumption an outward pointing vector field along $\partial X$, and $g$ is by definition the unit outward pointing vector field along $\partial X$. These two vector fields therefore take values in the same component of the open half space $T_p X - T_p \partial X$. So you can define a straight line homotopy between $v$ and $g$ which is never zero. You can then divide each vector in this homotopy by its norm to get a homotopy between $\bar v$ and $g$ taking values in $S^{m-1}$.