= Solution
Let $D$ be the ambient <Levi-Civita connection> and $\nabla$ the one for the <induced metric> on $M$. The <Gauss formula> is
$$
D_XY=\nabla_XY+II(X,Y),
$$
where the <second fundamental form> $II(X,Y)=(D_XY)^\perp$ is normal. The tangential part of $D$ is <metric-compatible> and <torsion-free>, so the <existence and uniqueness of the Levi-Civita connection> identifies it with $\nabla$.
For a normal field $\xi$, the <Weingarten formula> defines the <shape operator in a normal direction> $A_\xi$ and the <normal connection> by
$$
D_X\xi=-A_\xi X+\nabla_X^\perp\xi.
$$
Differentiate $\langle\xi,Y\rangle=0$ using metric compatibility to get
$$
\langle A_\xi X,Y\rangle=\langle II(X,Y),\xi\rangle.
$$
The <symmetry of the second fundamental form> follows from the ambient torsion-free condition:
$$
II(X,Y)-II(Y,X)=\bigl(D_XY-D_YX-[X,Y]\bigr)^\perp=0,
$$
since the <Lie bracket of vector fields> tangent to $M$ is tangent. Therefore
$$
\boxed{\langle A_\xi X,Y\rangle=\langle X,A_\xi Y\rangle},
$$
so every normal-direction <shape operator> is <self-adjoint>.
The paper uses $R(X,Y)=D_{[X,Y]}-[D_X,D_Y]$, so use the <Gauss equation with reversed curvature convention>. Applying the <Gauss formula> and <Weingarten formula> twice gives
$$
(D_XD_YZ)^\top=\nabla_X\nabla_YZ-A_{II(Y,Z)}X.
$$
Subtracting in the stated curvature order yields
$$
(R^V(X,Y)Z)^\top=R^M(X,Y)Z+A_{II(Y,Z)}X-A_{II(X,Z)}Y.
$$
Pair with a tangent field $W$ and use the defining relation for the <shape operator in a normal direction>:
$$
\boxed{\langle R^M(X,Y)Z,W\rangle
=\langle R^V(X,Y)Z,W\rangle
-\langle II(X,W),II(Y,Z)\rangle
+\langle II(Y,W),II(X,Z)\rangle}.
$$
This is the <Gauss equation> in the supplied sign convention. If one defines curvature by $[D_X,D_Y]-D_{[X,Y]}$ instead, the two quadratic terms have the opposite signs.
For an <orthonormal frame> $X,Y$ on a surface, positive <Gaussian curvature> with the convention here is $K=\langle R^M(X,Y)X,Y\rangle$. If the ambient space is Euclidean and $\xi$ is a unit normal, the equation becomes
$$
K=\langle II(X,X),II(Y,Y)\rangle-\|II(X,Y)\|^2
=\boxed{\det A_\xi}.
$$
The <Koszul formula> determines $\nabla$ from the <induced metric>, and hence determines its <Riemann curvature tensor> and $K$. Therefore the <Gaussian curvature> depends only on that metric, even though its expression as a product of <principal curvatures> appears extrinsic. This is <Theorema Egregium>.
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