The Curvature Tensor I present three different ways I've seen the Riemann curvature written:


*

*$R(X,Y)Z=D_XD_YZ-D_YD_XZ-D_{[X,Y]}Z$

*$R(e_c,e_d)e_b=D_{e_c}D_{e_d}e_b-D_{e_d}D_{e_c}e_b-D_{[e_c,e_d]}e_b$.

*$R^{\rho}_{\space \space\sigma \mu \nu}=dx^{\rho}(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})$
However I'm not told much more and I'm struggling to make sense of when we would need to use one or the other.
I'm assuming the $e_a$ in 2. are basis vectors, e.g. $e_a=\frac{\partial}{\partial x^1}$? So the second is just the curvature applied to basis vectors? But why would we want to do that? Also, doesn't $[e_c,e_d]=0$?
And with 3. I'm just completely lost. How do you relate $R(X,Y)Z$ and $R^{\rho}_{\space \space \sigma \mu \nu}$?
(This is quite an open question on the understanding of what's the intrinsic difference between these three and when we would need to use each in simple calculations. Or even a question about coordinate expressions of tensors in general.)
 A: There is probably someone more qualified to answer this, but I will take a crack at it.
Regarding 1.):
They are viewing $R : \Gamma\left(M\right) \times \Gamma\left(M\right) \times \Gamma\left(M\right) \to \Gamma \left(M\right)$, where $\Gamma\left(M\right)$ is the set of $C^{\infty}$ vector fields on the Riemannian manifold $M$.  This is very much an invariant definition of $R$ as it makes no appeal to coordinates, but rather only to objects that are invariantly associated to $M$ and its Riemannian metric: the Lie bracket of vector fields and the covariant derivative/connection.
The key observation here (in my opinion) is that $R$ is multilinear on the level of $C^{\infty}$ functions. I believe that most people refer to this as tensorial and it tells us that $R$ is induced by a $(3, 1)$ tensor (As a side note,  applying  the so called musical isomorphisms coming from the metric, one can obtain a corresponding $(4, 0)$ tensor field.)
Regarding 2.):
The answer is "Yes, this is evaluation of $R$ on the triple of vector fields $e_{a}, e_b, e_c$, where (presumably) $e_1$, $e_2, \ldots e_n$ constitue a frame field on $M$.  The key observation (and the connection with part 1.) and part 3.)) is the following: $R\left(e_a, e_b\right)e_c$ is a vector field on $M$.  This means that we can express  $R \left(e_a, e_b\right)e_c$ in terms if the frame field $e_{1}, \ldots e_{n}$ as
$$R \left(e_a, e_b\right)e_c = A^{i}e_{i},$$ 
where the $A^{i}$ are $C^{\infty}$ functions on $M$. (Note that we are making use of  the summation convention on repeated indices.) This leads us naturally to the connection  between 2.) and 1.),3.) . . . 
Regarding 3.):
Since $R : \Gamma\left(M\right) \times \Gamma\left(M\right) \times \Gamma\left(M\right) \to \Gamma \left(M\right)$, then in a coordinate frame field, we obtain the following 
$$R \left(\partial_a, \partial_b\right)\partial_c = A^{i}\partial_{i}.$$
Note that I have replaced the frame field $e_1, e_2, \ldots , e_n$ with the coordinate frame field $\partial_1, \partial_2, \ldots \partial_n$ and that while the form of the expression is the same, the coefficient functions $A^{i}$ depend on the frame used.
In order to find the coefficient functions $A^{i}$, we need to pair the vector field $R \left(\partial_a, \partial_b\right)\partial_c$  with the dual basis elements $dx^{\rho}$.  By doing so, we obtain
\begin{align*}
dx^{\rho}\left(R \left(\partial_a, \partial_b\right)\partial_c\right) &=  dx^{\rho}\left(A^{i}\partial_{i}\right)\\
&= A^{i} dx^{\rho}\left(\partial_{i}\right)\\
&=A^{i}\delta^{\rho}_{i}\\
&= A^{\rho}.
\end{align*}
Thus, the vector field $R \left(\partial_a, \partial_b\right)\partial_c$ is expressed with respect to the frame field as
$$ R \left(\partial_a, \partial_b\right)\partial_c = dx^{\rho}\left(R \left(\partial_a, \partial_b\right)\partial_c\right)\,\partial_{\rho}.$$
But this brings us full circle and expresses $R$ as a (3, 1) tensor field with respect to the coordinate frame field as
$$R^{\rho}_{a b c} dx^{a} \otimes dx^{b} \otimes dx^{c} \otimes \partial_{\rho}$$
where $R^{\rho}_{a b c}$ is the $\rho^{th}$ component of the vector field $R\left(\partial_a, \partial_b\right) \partial c$ with respect to the the coordinate frame.
Finally, regarding the issue of how to relate $R(X, Y)Z$ with $R^{\rho}_{a b c}$, we will express $X, Y$ and $Z$ in coordinates relative to the frame field and then rely on the multi-linearity of $R$ over $C$^{\infty}$ functions.  Observe that we have
\begin{align*}
X &= X^{i}\partial_{i}\\
Y &= Y^{j}\partial_{j}\\
Z &= Z^{k}\partial_{k},
\end{align*}
where indices on the coefficient functions $X^{i}, Y^{j}, Z^{k}$run from 1 to n.  Evaluating $R(X, Y)Z$, we obtain
\begin{align*}
R(X, Y)Z &= R\left(X^{i}\partial_{i}, Y^{j}\partial_{j}\right)Z^{k}\partial_{k} &\hspace{.5in} \textrm{(Sub)}\\
&= X^{i}Y^{j}Z^{k}R\left(\partial_{i}, \partial_{j}\right)\partial_{k} &\hspace{.5in} \textrm{(Mult-linearity)}\\
&= X^{i}Y^{j}Z^{k}R^{\rho}_{ijk}\partial_{\rho} &\hspace{.5in} \textrm{(Defn. of $R^{\rho}_{ijk}$)}
\end{align*}
Note: In the discussion of 3.), one could equally well use any frame field $e_{1}, \ldots , e_{n}$ and its corresponding dual frame field $\theta^{1}, \ldots, \theta^{n}$.
