The radii of curvature $R_1$ and $R_2$ at point $A$ are $R_1=AM$ ($M$ is the center of curvature of the meridian curve in the plane of the figure) and $R_2=AN$ (in the perpendicular plane). $\vec{n}$ is the normal to the surface at point $A$. x is the symmetry axis.
My question is whether $N$ must lies on the symmetry axis.Is the length of $AN$ equal to the radius of curvature parellel to the latitude? If not, is there any condition we need to add in order to make sure that the point $N$ lies on the symmetry axis?
First, let's define a surface of revolution with $z$ as the symmetry axis: $$f(t,\varphi) = \left ( r(t) \cos \varphi, \; r(t) \sin \varphi, \; h(t) \right )$$ For simplicity of notation, let's use $$r = r(t), \quad \dot{r} = \frac{d r(t)}{d t}, \quad \ddot{r} = \frac{d^2 r(t)}{d t^2} \\ h = h(t), \quad \dot{h} = \frac{d h(t)}{d t}, \quad \ddot{h} = \frac{d^2 h(t)}{d t^2}$$ The surface normal unit vector is $$\vec{n}(t, \varphi) = \left ( \frac{-\dot{h}\cos\varphi}{\sqrt{{\dot{r}}^2 + {\dot{h}}^2}}, \; \frac{-\dot{h}\sin\varphi}{\sqrt{{\dot{r}}^2 + {\dot{h}}^2}}, \; \frac{\dot{r}}{\sqrt{{\dot{r}}^2 + {\dot{h}}^2}} \right )$$ If $h\ne0$, and we have a point $\vec{p}(t, \varphi) = \left ( r\cos\varphi, r\sin\varphi, h \right)$ on the surface, and the unit normal $\vec{n}(t, \varphi)$ at that point, their slopes on the $xy$ plane are the same. This means that a line extending the unit normal will always intersect the symmetry axis ($z$ axis), as long as $h\ne0$. (If $h = 0$, then the surface normal is parallel to the symmetry axis.)
The principal curvatures -- maximum and minimum curvature at a given point $(t,\varphi)$ -- are $$\begin{cases} \kappa_1 = \frac{\dot{r}\ddot{h} - \ddot{r}\dot{h}}{({\dot{r}}^2+{\dot{h}}^2)^{3/2}} \\ \kappa_2 = \frac{\dot{h}}{r \sqrt{ {\dot{r}}^2 + {\dot{h}}^2 }} \end{cases}$$ and the Gaussian curvature is $$K = \kappa_1 \kappa_2 = - \frac{\ddot{r}}{r}$$
These hold as long as $r(t)$ and $h(t)$ are twice differentiable.
Radius of curvature is defined as the inverse of curvature. In that sense, the principal radii of curvature are $$\begin{cases} R_1 = \frac{1}{\kappa_1} = \frac{({\dot{r}}^2+{\dot{h}}^2)^{3/2}}{\dot{r}\ddot{h} - \ddot{r}\dot{h}} \\ R_2 = \frac{1}{\kappa_2} = \frac{r \sqrt{ {\dot{r}}^2 + {\dot{h}}^2 }}{\dot{h}} \end{cases}$$ In three dimensions, the centers of their respective circles are always on the line extending the surface normal. (They do differ in their orientation, rotation of the respective circle around the surface normal.)
Alternatively, we can think of the radii as the radii of two spheres, one defining the minimum curvature at that point on the surface, and the other the maximum curvature. The centers of the spheres are still on the line extending the surface normal.
If we look at the original illustration,
then, given that this is a surface of revolution, a line passing through $A$ and $M$, or through $A'$ and $M'$, will always pass through the symmetry axis.
In all honesty, I fail to see any relevance in the location of $M$, really. Since 3D surfaces have two principal curvatures at each point on the surface, using a single point to describe (a part of) it seems.. counterintuitive to me. It is much easier to use the surface normal and curvature parameters, in my opinion.