How do I show that the square of the gaussian curvature is the sum of the squares of the normal and geodesic curvatures other than the one shown in page 38 of http://www.maths.lancs.ac.uk/~belton/www/notes/geom_notes.pdf? That is, $$\kappa^2=\kappa_n^2 + \kappa_g^2$$
Is there another way?
The shape of a surface influences the curvature of curves on the surface.Let $S \subset \mathbb{R}^n$ a regular surface and $\alpha:(a, b) \rightarrow S$ a unit speed curve. If, for instance, $\alpha$ lies on a plane or a cylinder,
it can have zero curvature everywhere. This is not possible for curves on a sphere since no segment of a straight line can lie on a sphere. Thus the natural way to investigate how much a $S$ curves, consists in looking at the curvature of curves on the surface. In order to make everything crystal clar, please consider the picture below.
The geodesic and normal curvatures of $\alpha$ are related to the Gaussian curvature by the formula you shown above. Now by remembering that $\alpha'$ is a unit vector and it is tangent to $S$ we can conclude that $\alpha'$ is perpendiculat to the norlal $N$ of $S$.
Thus we have that $\alpha'$, $N$, and $N \times \alpha^{'}$ are mutually perpendicular unit vectors. Since $\alpha'$ is the unit speed, $\alpha''$ is perpendicular to $\alpha'$ and then it can be written as a linear combination of $N$ and $N \times \alpha^{'}$.
In other words we have:
\begin{equation}
\alpha^{''}(s)=\kappa_n N + k_g N \times \alpha^{'}(s)
\end{equation}
Where
\begin{equation}
\kappa_n =\alpha^{''} \cdot N
\end{equation}
and
\begin{equation}
\kappa_g =\alpha^{''} \cdot (N \times \alpha^{'})
\end{equation}
Now:
\begin{equation}
\kappa_\alpha(s)^2=||\alpha^{''}(s)||^2=||\alpha^{''}(s)^{\parallel}||^2+||\alpha^{''}(s)^{\perp}||^2
\end{equation}
i.e. the parallel and perpedicular components of the acceleration of $\alpha(s)$. By definition $\alpha^{''}(s)^{\parallel}$ is the component tangent to $S$ and $\alpha^{''}(s)^{\perp}$ s the component parallel to a normal vector to the surface.
The $\kappa$ in your question is not the Gaussian curvature. The Gaussian curvature is the product of two normal curvatures. The $\kappa$ in your question is the scalar curvature, or the magnitude of the curvature vector $\mathbf{k}$ at a regular point P of the curve $C$ that lays on a surface $S$.
