What is $\nabla X$ in Riemannian geometry? What exactly is $\nabla X$, for a vector field $X$? Can we write it in coordinates?

I'm familiar with $\nabla f$, and also $\nabla_Y X$, where both $X$ and $Y$ are vector fields. However, what would $\nabla X$ be? Should we think of it as a map $TM\to TM$?
I imagine this is in the context of differential geometry (rather than vector calculus, where I would not know what it stands for). Then it is very simply related to $\nabla_YX $ which you say you are familiar with:
\begin{equation}
\nabla X (Y) = \nabla _Y X.
\end{equation}
While the covariant derivative preserves a tensor field rank (that is, the covariant derivative of a vector is a vector, of a 1-form is a 1-form, and so on), the action of $\nabla$ itself increases the covariant (differential form) degree by one. Hence we can characterise it by saying how it acts on vector fields as above.
"nabla", ∇, can be interpreted as the "vector operator", $\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}$.

There are three different kinds of vector multiplication- scalar product (between a scalar and a vector), dot product, and cross product (between two vectors)- so there are three different ways to use nabla:

"gradient": One takes the "grad" of a scalar valued function of x, y, and z and the result is a vector valued function.

$\nabla f= \left(\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}\right) f(x, y, z)= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$.  

That is the vector pointing in the direction of fastest increase of f.  It's length is the rate of increase in that direction.

"divergence"  One takes the "dot product" with a vector valued function and the result is a scalar.

$\nabla\cdot f\vec{i}+ g\vec{j}+ h\vec{k}= $$\left(\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}\right)\cdot \left(f\vec{i}+ g\vec{j}+ h\vec{k}\right)=$$ \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial g}{\partial y}\vec{j}+ \frac{\partial h}{\partial z}\vec{k}$.

If we think of the vector function as the velocity vector of a fluid at each point, that measures, roughly, how fast the fluid "spreads out".

"curl":  One takes the "cross product" with a vector valued function and the result is another vector value function.

$\nabla \times \left(f\vec{i}+ g\vec{j}+ h\vec{k}\right)= \left|\begin{array}{ccc} \vec{i} &amp; \vec{j} &amp; \vec{k} \\ \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} \\ f &amp; g &amp; h \end{array}\right|= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}- \left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$

If we think of the vector function as the velocity vector of a fluid at each point, that measures, roughly, the rotation of the fluid.

Also of great importance in physics is the "Laplacian",  the divergence of the gradient of a scalar valued function:

$\nabla^2 f= \nabla\cdot\left(\nabla f\right)= \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2}+ \frac{\partial^2 f}{\partial x^2}$

For some reason physicists tend to write that as "$\triangle f$"