This is problem 17.3 from Tu's Differential Geometry text. This post appears to answer my question, but I don't follow many of the explanations, including the construction and usage of $det$.
Problem
Let $M$ be a smooth, compact, oriented surface in $\mathbb{R^3}$ with $K$ the Gaussian curvature on $M$. If $v : M \to S^2 $ is the Gauss map, prove that
$$ v^* (vol_{S^2}) = K vol_M $$
Corrected Proof
Lemma 1. If $N$ is an $n$-dimensional smooth, oriented manifold, $(U, x^1, \ldots, x^n)$ a chart a $p \in N$ with onb $\{e_1, \ldots e_n \} =: e $ , then
$$ vol_N = det_e $$
Proof. If $\theta^1, \ldots \theta^n$ is dual to $e_1, \ldots e_n$, then for any $X_1, \ldots X_n$ in $T_p N$ with $X_j = \sum a^i_j e^i$
\begin{align}
vol_N (X_1, \ldots X_n) &= \theta^n \wedge \cdots \wedge \theta^n (X_1, \ldots X_n) \\
&= \sum_{\sigma \in S_n} sgn(\sigma) a^i_{\sigma(1)} \cdots a^i_{\sigma(n)} \\
&= det_e [ a^i_j ]
\end{align}
$\square$
Lemma 2. There exists $\{ e_1, e_2 \} =: e$ principal vectors at $p$ in $M$ such that $e$ is an onb for $M$, and $ \{ \kappa_1 e_1, \kappa_2 e_2 \} = \{ dv_p(e_1), dv_p(e_2) \} =: e'$ is an orthogonal basis for $S^2$.
We will use these facts: $dv_p = -L$ (the shape operator at $p$) (Problem 5.3), the $L$ has the principal vectors as eigenvectors and principal curvatures as eigenvalues (Prop 5.6), and $L$ is self-adjoint.
Note that if $\kappa_1 = \kappa_2$, all vectors in $T_p M$ are principal, so we can freely choose an onb. So assume they're unequal.
Then
\begin{align}
\kappa_1^2 \langle e_1, e_2 \rangle &= \langle -L (e_1), -L (e_2) \rangle \\
&= \langle dv_p (e_1), dv_p (e_2) \rangle \\
&= \kappa_2^2 \langle e_1, e_2 \rangle
\end{align}
Because the principal curvatures are not equal, we must have
$$\langle e_1, e_2 \rangle = 0 = \langle dv_p (e_1), dv_p (e_2) \rangle = \langle \kappa_1 e_1, \kappa_2 e_2 \rangle.$$
$\square$
Note that Lemma 2 implies that $e$ is an onb for both $T_p M$ and $T_p S^2$.
Now if $K(p) = det(J_v(p)) = 0$ then the claim obviously holds because $v^* vol_{S^2} = 0 = K(p) vol_M$. So assume $K(p) = det(J_V(p)) \neq 0$.
Applying Lemma 2, Choose onb $e$ for $M$ (which is also onb for $S^2$). Let $\Theta^1 \wedge \Theta^2$ be the volume form for $S^2$ (with $\Theta^i$ dual to $e_i$). Then for any $X_1, X_2$ vectors in $S^2$ with $X_j = \sum x^{i}_j e_i$.
\begin{align}
v^* (vol_{S^2})(X_1, X_2) &= \Theta^1 \wedge \Theta^2 (dv_p X_1, dv_p X_2) \\
&= \Theta^1(dv_p X_1) \Theta^2(dv_p X_2) – \Theta^1(dv_p X_2) \Theta^2(dv_p X_1) \\
\text{(Lemma 2)} &= (\kappa_1 x^1_1) (\kappa_2 x^2_2) – (\kappa_1 x^1_2) (\kappa_2 x^2_1) \\
&= K det_e(X_1, X_2) \\
\text{(Lemma 1)} &= K vol_M
\end{align}
$\square$
Update
Special thanks to @cbishop for providing the key insight (Prop 8.2.1) that helped me solve the problem!
Best Answer
Corollary 8.2.2 of Pressley's book "Elementary Differential Geometry" states that there is an orthonormal basis of the tangent plane of a surface consisting of the principal vectors. I believe that this fact plus multi-linearity should complete your proof.
I had the wrong author above, that is now fixed. A screenshot of the proof of the corollary is here:
Here is the statement of A.0.3 from Pressley:
So since the Weingarten map is self-adjoint, it is diagonalizable and the eigenvalues are the principal curvatures. It appears that this is covered in Tu's book in Section 5.3 up through Corollary 5.7.