Let’s go general to special. Suppose $(M,g)$ is a Riemannian submanifold of $(\widetilde{M},\widetilde{g})$, and let $\alpha:TM\oplus TM\to (TM)^{\perp}$ be the (vector) second fundamental form. Then, for each point $p\in M$ and linearly independent tangent vectors $x,y\in T_pM$, Gauss’ equation tells us
\begin{align}
A_p(x,y)&=\widetilde{A}_p(x,y)+\frac{\langle\alpha(x,x),\alpha(y,y)\rangle-\|\alpha(x,y)\|^2}{\|x\|^2\|y\|^2-\langle x,y\rangle^2},
\end{align}
where $A_p(x,y)$ is the sectional curvature of the tangent plane in $T_pM$ spanned by $x,y$ as measured in $(M,g)$, and $\widetilde{A}_p(x,y)$ is the sectional curvature measured by $(\widetilde{M},\widetilde{g})$. So, this equation tells you that the sectional curvature of the submanifold (which is intrinsic to the submanifold) equals the sectional curvature of the ambient manifold (intrinsic to the ambient manifold) + some stuff involving the second fundamental form (i.e extrinsic to the submanifold).
Now, suppose $\dim \widetilde{M}=n\geq 3$ and $M$ is a hypersurface. Fixing a unit normal $\nu$ for $T_pM$, we have that the shape operator/Weingarten map $S_{\nu}: T_pM\to T_pM$ is related to the second fundamental form $\alpha$ by $\alpha(x,y)=\langle S_{\nu}(x),y\rangle\nu$. With this, the above equation becomes
\begin{align}
A_p(x,y)&=\widetilde{A}_p(x,y)+\frac{\langle S_{\nu}(x),x\rangle\langle S_{\nu}(y),y\rangle-\langle S_{\nu}(x),y\rangle^2}{\|x\|^2\|y\|^2-\langle x,y\rangle^2},
\end{align}
Symmetry of $\alpha$ implies $S_{\nu}$ is self-adjoint, so by the spectral theorem, we can find an orthonormal basis of eigenvectors $\{e_1,\dots, e_{n-1}\}$ of $S_{\nu}$ for $T_pM$, say with corresponding eigenvalues $\lambda_1,\dots,\lambda_{n-1}$. Then, for each $1\leq i<j\leq n-1$, we have that
\begin{align}
A_p(e_i,e_j)&=\widetilde{A}_p(e_i,e_j)+\lambda_i\lambda_j,
\end{align}
or rearranging, $\lambda_i\lambda_j=A_p(e_i,e_j)-\widetilde{A}_p(e_i,e_j)$
. Now, multiply over all possible pairs $i,j$ such that $1\leq i<j\leq n$. Then, we get
\begin{align}
(\lambda_1\cdots\lambda_{n-1})^{n-2}&=\prod_{1\leq i<j\leq n-1}\left[A_p(e_i,e_j)-\widetilde{A}_p(e_i,e_j)\right].
\end{align}
On the left this is precisely $(\det S_{\nu})^{n-2}$, and on the right, we have a bunch of products of the difference in sectional curvatures. Therefore, depending on the parity of the dimension $n$ of the ambient manifold, we can solve for $\det S_{\nu}$:
\begin{align}
\begin{cases}
\det S_{\nu}=\left(\prod_{1\leq i<j\leq n-1}\left[A_p(e_i,e_j)-\widetilde{A}_p(e_i,e_j)\right]\right)^{\frac{1}{n-2}}&,\quad\text{if $n$ odd}\\
|\det S_{\nu}|= \left(\prod_{1\leq i<j\leq n-1}\left[A_p(e_i,e_j)-\widetilde{A}_p(e_i,e_j)\right]\right)^{\frac{1}{n-2}}&,\quad\text{if $n$ even}
\end{cases}
\end{align}
In particular,
- if $n\geq 3$ is odd, and the ambient space $\widetilde{M}$ is flat then its sectional curvatures all vanish, and so we have expressed $\det S_{\nu}$ completely in terms of the sectional curvatures $A$ of the submanifold (which is completely intrinsic to the submanifold), thereby proving that $\det S_{\nu}$ (which is calculated in an extrinsic fashion) actually depends only on the geometry of the submanifold. Also, reading the formula right-to-left, we see that it does not matter which orthonormal basis $\{e_1,\dots, e_{n-1}\}$ one chooses to compute the sectional curvatures; the resulting product is independent of this choice.
- if $n\geq 4$ is even, and the ambient space is flat, then it is $|\det S_{\nu}|$ which is intrinsic to the
That this sign ambiguity must be present is obvious because the choice of normal $\nu$ is only determined up to sign, so replacing $\nu$ by $-\nu$ changes $\det S_{\nu}$ to $\det S_{-\nu}=\det(-S_{\nu})=(-1)^{n-1}\det S_{\nu}$, i.e it stays the same if $n$ is odd and flips sign if $n$ is even.
Since $\det S_{\nu}$ is originally how Gauss defined his curvature, we get the result that the Gaussian curvature is intrinsic to the submanifold. In particular, specializing to $n=3$ with flat ambient space, we see that there is only one term on the right:
\begin{align}
\det S_{\nu}=A_p(e_1,e_2)\equiv A_p,\tag{$*$}
\end{align}
where $A_p\in\Bbb{R}$ denotes the sectional curvature of the plane $T_pM$ (there is only one since $M$ is 2-dimensional).
A completely separate result is the following: for any $(n-1)$-dimensional Riemannian manifold $(M,g)$, the scalar curvature of $(M,g)$ is equal to the sum of the distinct sectional curvatures in any orthonormal basis $\{e_1,\dots, e_{n-1}\}$ for $T_pM$:
\begin{align}
R_p&=2\sum_{1\leq i<j\leq n-1}A_p(e_i,e_j).
\end{align}
This follows just by definition of the various things, and using orthonormality. In particular, specializing to $n-1=2$-dimensional manifold $M$, there is only one term on the right, so
\begin{align}
R_p&=2A_p.\tag{$**$}
\end{align}
So, if you now combine the two situations $(*)$ and $(**)$, we get that for a Riemannian 2-manifold $M$ embedded in a flat 3-dimensional space,
\begin{align}
\det S_{\nu}&=A_p=\frac{R_p}{2}.\tag{$***$}
\end{align}
It is this equation, $(***)$, which motivates the abstract intrinsic definition of Gaussian curvature, $K$, of a Riemannian 2-manifold $(M,g)$: we define $K:=\frac{R}{2}$, where $R$ is the scalar curvature. So, with this definition, the Gaussian curvature of a Riemannian 2-manifold coincides with the single sectional curvature $A$ of $M$, and when embedded into $\Bbb{R}^3$ it coincides with $\det S_{\nu}$, the product of the principal curvatures. So, I think you’re having an issue with what’s a definition vs what’s being proved.
Best Answer
In a course of a colleague, I found the following result that might be useful to you. The idea is to look at the tangent parts of $x_{uu}, x_{uv}, x_{vv}$, etc. Recall that the tangent part of the derivative of a tangent vector is the covariant derivative on the surface.
Proof: If $N$ is a normal unit vector field along $x$, then $$ \begin{align*} x_{vv}^T &= x_{vv}+(x_v\cdot N_v)N \\ x_{uv}^T &= x_{uv}+(x_u\cdot N_v)N. \end{align*} $$ Deriving gives $$ \begin{align*} (x_{vv}^T)_u &= x_{uvv}+(x_v\cdot N_v)_u N + (x_v\cdot N_v) N_u \\ (x_{uv}^T)_v &= x_{uvv}+(x_u\cdot N_v)_v N + (x_u\cdot N_v) N_v, \end{align*} $$ and thus $$ (x_{vv}^T)_u - (x_{uv}^T)_v=\bigl((x_v\cdot N_v)_u-(x_u\cdot N_v)_v\bigr)N +(x_v\cdot N_v)N_u - (x_u\cdot N_v)N_v. $$ Take the inproduct with $x_u$. We use the definition of the shape operator, the functions $E$, $F$, \ldots, $f$ and $g$ and the Gauss curvature $K$ to obtain $$ \begin{align*} \bigl((x_{vv}^T)_u - (x_{uv}^T)_v\bigr)\cdot x_u &= (x_v\cdot N_v)(N_u\cdot x_u)-(x_u\cdot N_v)(N_v\cdot x_u)\\ &= eg - f^2 = (K\circ x) (EG-F)^2. \end{align*}$$
To prove the Theorema Egregium, one should also use the following result: