= Solution
The <tangent vectors> of
$$
\sigma(u,v)=(u,v,f(u,v))
$$
are
$$
\sigma_u=(1,0,f_u),
\qquad
\sigma_v=(0,1,f_v).
$$
Hence the <first fundamental form> has coefficients
$$
E=1+f_u^2,
\qquad
F=f_uf_v,
\qquad
G=1+f_v^2.
$$
Put
$$
W=\sqrt{1+f_u^2+f_v^2}.
$$
The upward <unit normal> is $N=(-f_u,-f_v,1)/W$, so the <second fundamental form> has coefficients
$$
e=\frac{f_{uu}}W,
\qquad
f_{II}=\frac{f_{uv}}W,
\qquad
g=\frac{f_{vv}}W.
$$
Thus the two forms are
$$
I=E\,du^2+2F\,du\,dv+G\,dv^2,
\qquad
II=e\,du^2+2f_{II}\,du\,dv+g\,dv^2.
$$
The <Gaussian curvature> is
$$
K=\frac{eg-f_{II}^2}{EG-F^2}.
$$
Since $EG-F^2=W^2$, the graph formula is
$$
\boxed{
K=\frac{f_{uu}f_{vv}-f_{uv}^2}
{(1+f_u^2+f_v^2)^2}
},
$$
as in <Gaussian curvature of a graph surface>.
For the final claim, fix a point of $\gamma$ and make a <rigid motion> taking $P$ to the plane $z=0$. The common tangent plane is horizontal, so locally $\Sigma$ is the graph $z=f(u,v)$. Along the projected curve $c(s)$ of tangency,
$$
f(c(s))=0,
\qquad
\nabla f(c(s))=0.
$$
Differentiating the second identity gives
$$
\operatorname{Hess}f(c(s))\,c'(s)=0.
$$
Because $c$ is a <smooth curve>, $c'(s)\ne0$, so the Hessian is singular. Its <determinant> is zero, and the graph formula gives $K=0$ at every point of $\gamma$. This is <tangency to a plane along a curve forces zero Gaussian curvature>.
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