geometric meaning of the second fundamental form in a Riemannian manifold Let $(M^{n+1},g)$ be a Riemannian manifold and let $\Sigma^n$ be a hypersurface. Locally, using the Riemannian polar coordinates, we have the "out-pointing" vector field $\nu$.
The second fundamental form, h, of $\Sigma$ is symmetric so there exist $e_1, \ldots, e_n$ orthonormal which diagonalise h. On the diagonal we have the principal curvatures $\lambda_1, \ldots, \lambda_n$. 
I'm suppose to think that the principal curvatures are the "curvature" of $\Sigma$ intersected with an hyperplane generated by $(n-1)$ vectors chosen in ${e_1, \ldots, e_n}$ and $\nu$. Is it correct? How can I prove it?
 A: Partial answer

I'm suppose to think that the principal curvatures are the "curvature" of $\Sigma$ intersected with an hyperplane generated by $(n-1)$ vectors chosen in ${e_1, \ldots, e_n}$ and $\nu$. Is it correct? How can I prove it?

That's correct, although "generated by" is a little tricky. 
It helps to first think of the case where $M$ is $3$-space, and $\Sigma$ is an ordinary surface, and at some point $P$, the normal to $\Sigma$ is $\nu$. There are (in this case) two principal curvatures, in directions $e_1, e_2$, and they correspond to the intersection of $M$ with the planes through $P$ spanned by (1) $e_1, \nu$, and (2) $e_2, \nu$. 
In fact, consider the set of all planes containing the point $P$ and direction $\nu$. They can be distinguished by an angle $0 \le \theta < \pi$ from some chosen plane in the set. The intersection of plane $X_\theta$ with $\Sigma$ is a curve, and the curvature of this curve is some number. It takes a little work to show that $k(\theta)$ has exactly one max and one min on the interval $0 \le \theta < \pi$, and these are the principal curvatures. 
But to return to the original statement: "There are (in this case) two principal curvatures, in directions $e_1, e_2$, and they correspond to the intersection of $M$ with the planes through $P$ spanned by (1) $e_1, \nu$, and (2) $e_2, \nu$." How to generalize this? 
One thought is that in $n$ dimensions, there are just $n$ cases: curvatures in 2-planes generated by $e_i, \nu$, where $i = 1 \ldots n$.
The other is that rather than seeing $e_1, \nu$ as "pairing one tangent vector with $\nu$, you can see it as "pairing up all but one tangent vector with $\nu$," and that's the claim you're trying to work with. So for a 3-manifold $\Sigma$, you'd be looking at planes spanned by $e_2, e_3, \nu$, by $e_1, e_3, \nu$, and by $e_1, e_2, \nu$. 
As for "how to prove it," that depends on what definition you have for the second fundamental form, so I'm going to stop here. I recommend that you carefully work through the proof in the case for a surface in 3-space (using whatever definition you ARE using), and try to see whether any parts of it fail to generalize. 
