= Solution
At $P=\sigma(u,v)$, regularity makes $\sigma_u,\sigma_v$ linearly independent. The <tangent space> is $T_PU=\operatorname{span}\{\sigma_u,\sigma_v\}$, and an oriented unit <normal vector> is
$$
\boldsymbol n=\frac{\sigma_u\times\sigma_v}{|\sigma_u\times\sigma_v|}.
$$
Reversing its sign chooses the opposite orientation. The <first fundamental form> is the squared length of a tangent displacement:
$$
|\sigma_u\,du+\sigma_v\,dv|^2=E\,du^2+2F\,du\,dv+G\,dv^2,
$$
where $E=\sigma_u\cdot\sigma_u$, $F=\sigma_u\cdot\sigma_v$, $G=\sigma_v\cdot\sigma_v$. Its determinant $EG-F^2=|\sigma_u\times\sigma_v|^2$ is positive.
Use $N_{\rm II}=\sigma_{vv}\cdot\boldsymbol n$ to distinguish the scalar coefficient of the <second fundamental form> from the <normal vector>. Since $\boldsymbol n\cdot\boldsymbol n=1$, its derivatives are perpendicular to it and hence tangent. Write $\boldsymbol n_u=a\sigma_u+b\sigma_v$ and $\boldsymbol n_v=c\sigma_u+d\sigma_v$. Differentiating $\boldsymbol n\cdot\sigma_u=\boldsymbol n\cdot\sigma_v=0$ gives
$$
\boldsymbol n_u\cdot\sigma_u=-L,\quad \boldsymbol n_u\cdot\sigma_v=-M,\quad
\boldsymbol n_v\cdot\sigma_u=-M,\quad \boldsymbol n_v\cdot\sigma_v=-N_{\rm II}.
$$
Substitution of the tangent expansions proves
$$
\boxed{-\begin{pmatrix}L&M\\M&N_{\rm II}\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}E&F\\F&G\end{pmatrix}.}
$$
The <shape operator> is $-d\boldsymbol n$; its <matrix> in the parametrized tangent basis is the negative transpose of the coefficient <matrix> shown. Its determinant, the product of the principal curvatures, is therefore
$$
\boxed{ad-bc=\frac{LN_{\rm II}-M^2}{EG-F^2}=K,}
$$
the <Gaussian curvature>. Transposition and the two minus signs do not change this determinant.
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