Why is the square of curvature of a curve equal to the square of curvature of a normal plus the square of geodesic curvature? I don't know what the normal curvature is. Is it the curvature of the normal direction of the surface? Or the curvature in the normal direction of the curve? Can we use graphics to explain the relationship between these three curvatures?
 A:  $(k_n, k_g)$
Yes normal curvature is in the osculating plane fully described by principal curvatures $k_1,k_2$  with Euler's relation.
Reference to figure at bottom of page 5.
There are two distinct mutually perpendicular components $(k_n, k_g)$ in normal and tangential planes respectively.
When surface normal and arc line normal of stand alone Frenet-Serret do not coincide we have a non-geodesic situation. The rate at which the angle deviates is  $ k_g.$ 
If it vanishes we are talking about a geodesic line.
In a globe the Equator is free of geodesic curvature. Other  parallels are latitudes which are not geodesics. At poles there is maximum  $k_g.$ The curvatures or accelerations in a dynamic situation e.g. banking of curves can be compared.
$ \tan \gamma= \dfrac{  k_g}{ k_n} $ describes the degree of slip-off and angle $\gamma$ lies in the plane spanned by the normal N and the binormal B.

In a Loxodrome (rhumb line) for example, the tangent starts as a geodesic at Equator and $k_g$ increases indefinitely while approaching towards either Pole along shrinking spiral paths.
