On pages 118-119 of Spivak's Calculus on Manifolds he introduces the idea of consistent orientations by stating
It is often necessary to choose an orientation $\mu_x$ for each tangent space $M_x$ of a manifold $M \subset \mathbb R^n$. Such choices are called consistent provided that for every coordinate system $f:W\rightarrow M \subset \mathbb{R}^n$ and $a,b\in W$ the relation $$[f_*((e_1)_a),\ldots,f_*(e_k)_a)]=\mu_{f(a)}$$ holds if and only if $$[f_*((e_1)_b),\ldots,f_*(e_k)_b)]=\mu_{f(b)}.$$ Suppose orientations $\mu_x$ have been chosen consistently. If $f:W\rightarrow M \subset \mathbb{R}^n$ is a coordinate system such that $$[f_*((e_1)_b),\ldots,f_*((e_k)_b)]=\mu_{f(a)}$$ for one, and hence every $a\in W$, then $f$ is called orientation-preserving.
He then goes on to say
If $f$ and $g$ are orientation-preserving and $x=f(a)=g(b)$, then the relation $$[f_*((e_1)_a),\ldots,f_*((e_k)_a)]=\mu_x=[g_*((e_1)_b),\ldots,g_*((e_k)_b)]$$ implies that $$[(g^{-1}\circ f)_*((e_1)_a,\ldots,(e_k)_a)]=[(e_1)_b,\ldots,(e_k)_b],$$ so that $\det\,(g^{-1}\circ f)'>0$, an important fact to remember.
It is this last part that slightly confuses me. It seems intuitively clear, but a rigorous proof of this has eluded me. I have a few questions regarding the matter. Why is Spivak able to apply $(g^{-1})_*$ and keep the bases in the same orientation class? How do we know that $(g^{-1})_*$ applied to the vectors $f_*((e_i)_a)$ and $g_*((e_i)_b)$ for $i=1,\ldots,k$ doesn't impact the orientations in different ways (i.e. reverse one and preserve the other)? Also, if it is true that it doesn't influence the orientations differently, then does this same fact apply to $g_*$ and other coordinate systems for that matter? I believe that $(g^{-1})_*g_*=I_k$ (the $k\times k$ identity matrix, not sure if helpful). Lastly, I know we call $f,g$ orientation preserving, but does $[f_*((e_1)_a),\ldots,f_*((e_k)_a)]=\mu_{f(a)}$ imply $[f_*((v_1)_a),\ldots,f_*((v_k)_a)]=\mu_{f(a)}$ for any basis vectors $v_i$ with the same orientation as the standard basis on $\mathbb{R}^k$?
Note: The square brackets are used by Spivak to represent the equivalence class for the orientation of a given basis. The orientation to which a basis $\{v_1,\ldots,v_k\}$ belongs is denoted $$[v_1,\ldots,v_k]$$ and the other orientation is denoted $$-[v_1,\ldots,v_k].$$ If $v_1,\ldots,v_k$ and $w_1,\ldots,w_k$ are two bases and $A=(a_{ij})$ is defined by $w_i=\sum a_{ij}v_j$, then $v_1,\ldots,v_k$ and $w_1,\ldots,w_k$ are in the same orientation class if and only if $\det(A)>0$. Also, $e_1,\ldots,e_k$ denote the standard basis vectors in $\mathbb{R}^k$.
The general linear algebra fact that you need here is this:
To prove this, let $A$ be the change of basis matrix from the $v$'s to the $w$'s, i.e. $w_i = \sum_j a_{ij} v_j$. From the assumption that $[v_1,\ldots,v_k]=[w_1,\ldots,w_k]$ it follows that $\det(A)>0$. Using the properties of linear transformations it follows that $$T(w_i) = T\left(\sum_j a_{ij} v_j\right) = \sum_j a_{ij} T(v_j) $$ Therefore $A$ is also the change of basis matrix from the $T(v)$'s to the $T(w)$'s. Since $\det(A)>0$, it follows that $[T(v_1),...,T(v_k)]=[T(w_1),\ldots,T(w_k)]$.
One can apply this in your setting using $T = (g^{-1})_*$, $V_1 = T_x M$, and $V_2 = T_b W = T_b \mathbb R^k = \mathbb R^k$, to derive the implication in your second quotation from Spivak.