Covariant derivative of vector field along itself: $\nabla_X X$
Solution 1:
As Mike Miller says, vector fields with $\nabla_XX=0$ are very special. For such a vector field, every integral curve is a geodesic. In the plane, for example, what does such a vector field look like? The direction of the vector field has to be constant, and the magnitude can only change in the direction perpendicular to $X$. In other words, $X$ looks like a bunch of parallel stripes, with each stripe having constant magnitude, such as $X(x,y) = (y^2,0).$
There are several intuitive physical interpretations of $X$:
Consider the case where you are on a submanifold of $\mathbb{R}^3$. Put a particle at a point $p$ on the manifold and give it initial velocity $X(p)$. Let the particle travel inertially over the manifold, constraining it to stay on the manifold and not "lift off" into ambient space, i.e. at every point in time, apply just enough acceleration in the normal direction to the manifold to keep the particle's velocity tangent to the manifold. (Think of a magnetic ball bearing, rolling over a sheet of steel in the shape of your manifold). Then the particle will travel along integral curves of $X$, that is its velocity at any time $t$ will be $X(p(t))$.
Cover the manifold in (infinitely compressible) fluid, and give the fluid initial velocity $X$. Now allow the fluid to flow for any amount of time $t$ without any forces acting on it. The fluid velocity at time $t$ will look exactly the same as at time $0$, $X(t)=X$.
When $\nabla_XX \neq 0$, the covariant derivative gives you the failure, at that point, of the vector field to have geodesic integral curves; in interpretation #1 above, for instance, it's the tangential force you must apply to the particle to make it follow the vector field with velocity $X(p(t))$. In interpretation #2, it gives you the negative time derivative of the fluid velocity at a given point (the acceleration felt by fluid particles at that point).