What does Ricci tensor do with two vectors? I have found it easier to understand the meaning of a particular tensor if I can find out what does it do if I cancel all its lower indices with vectors in short:


*

*$g_{ij} u^i v^j$: dot product of $\mathbf u$ and $\mathbf v$.

*$-\Gamma^k_{ij} u^i v^j$ : rate of change of the local coordinates of $\mathbf u$ when parallel transported in the direction of $\mathbf v$.

*$R^l_{ijk} s^i u^j v^k$ : form a parallelogram from $\mathbf u$ and $\mathbf v$ at the given point, parallel transport $\mathbf s$ around it, the resulting vector represents the change of $\mathbf s$ after completing the loop, thus indicating the curvature of the manifold near that the point.


Now doing the same with the Ricci tensor, so $R_{ij} u^i v^j$.
So you supply 2 vectors, and you get a number. What's the meaning of the vectors in this case? What's the meaning of the resulting number?
I read that if the 2 vectors are the same then it tells how does the volume of a ball moved along a geodesic changes. Just like the Christoffel symbols tell you what acceleration you will feel if you go straight (in coordinates) towards a given direction when you supply the same vector twice.
But I'd like to know the more general case when the two vectors isn't the same.
 A: The Ricci tensor obviously a tensor that accepts two vectors and outputs a number. This number represents in some sense the "average" sectional curvature at a given location on the manifold $M$. In GR, we usually use spacetime manifolds. 
In his book Riemannian Geometry, Manfredo Do Carmo states the following on page 97: 

Let $x = z_n$ be a unit vector in $T_pM$; we take an orthonormal basis $\lbrace z_1,z_2,...,z_{n-1}\rbrace$ of the hyperplane in $T_pM$ orthogonal to $x$ and consider the following averages: 
  \begin{equation}
\text{Ric}_p(x) = \frac{1}{n-1} \sum_i \langle R(x,z_i)x,z_i \rangle 
\end{equation}
  for $i = 1,2,...,n-1$

This is a hint of a more formal version of what I said in the first paragraph. Let me add a bit more. 
If $u, v$ are tangent vectors at a point $p \in M$, then
$$
   Ric(u,v) = R_{ab} u^a v^b$$
Here, $u^a, v^b$ is abstract index notation. Note, 
$Ric$ is symmetric, so one way of thinking about $Ric$ is to interpret $Ric(u,u)$ for all tangent vectors $u$ (since this determines $Ric$ in general), and it suffices then to just try to understand this when $||u|| = 1$. 
By definition $Ric$ is the trace of the Riemann curvature tensor on its first and last indices. If $ u \in T_pM$ choose an orthonormal basis $e_1, \dots, e_n$ for $T_pM$ with $e_1 = u$; then
$$
  Ric(u, u) = \sum_{i = 1}^n \langle R(e_i,u)u, e_i \rangle.
$$
(If you are only familiar with the abstract index notation, the thing on the right is given by
$$
  \langle R(u, v) x, y \rangle = g_{\tau d} R_{abc}^{\phantom{abc}\tau} u^a v^b x^c y^d, 
$$
So when $e_i, u$ are orthonormal, $\langle R(e_i, u)u, e_i \rangle$ gives the sectional curvature of the plane spanned by $e_i$ and $u$. In particular,
$$
  \frac{1}{n-1} Ric(u, u),
$$ 
when $||u|| = 1$, is an average of the sectional curvatures of 2-planes through $u$.
