The simplest way to explain the Christoffel symbol is to look at them in flat space. Normally, the laplacian of a scalar in three flat dimensions is:
$$\nabla^{a}\nabla_{a}\phi = \frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}$$
But, that isn't the case if I switch from the $(x,y,z)$ coordinate system to cylindrical coordinates $(r,\theta,z)$. Now, the laplacian becomes:
$$\nabla^{a}\nabla_{a}\phi=\frac{\partial^{2}\phi}{\partial r^{2}}+\frac{1}{r^{2}}\left(\frac{\partial^{2}\phi}{\partial \theta^{2}}\right)+\frac{\partial^{2}\phi}{\partial z^{2}}-\frac{1}{r}\left(\frac{\partial\phi}{\partial r}\right)$$
The most important thing to note is the last term above--you now have not only second derivatives of $\phi$, but you also now have a term involving a first derivative of $\phi$. This is precisely what a Christoffel symbol does. In general, the Laplacian operator is:
$$\nabla_{a}\nabla^{a}\phi = g^{ab}\partial_{a}\partial_{b}\phi - g^{ab}\Gamma_{ab}{}^{c}\partial_{c}\phi$$
In the case of cylindrical coordinates, what the extra term does is encode the fact that the coordinate system isn't homogenous into the derivative operator--surfaces at constant $r$ are much larger far from the origin than they are close to the origin. In the case of a curved space(time), what the Christoffel symbols do is explain the inhomogenities/curvature/whatever of the space(time) itself.
As far as the curvature tensors--they are contractions of each other. The Riemann tensor is simply an anticommutator of derivative operators--$R_{abc}{}^{d}\omega_{d} \equiv \nabla_{a}\nabla_{b}\omega_{c} - \nabla_{b}\nabla_{a} \omega_{c}$. It measures how parallel translation of a vector/one-form differs if you go in direction 1 and then direction 2 or in the opposite order. The Riemann tensor is an unwieldy thing to work with, however, having four indices. It turns out that it is antisymmetric on the first two and last two indices, however, so there is in fact only a single contraction (contraction=multiply by the metric tensor and sum over all indices) one can make on it, $g^{ab}R_{acbd}=R_{cd}$, and this defines the Ricci tensor. The Ricci scalar is just a further contraction of this, $R=g^{ab}R_{ab}$.
Now, due to Special Relativity, Einstein already knew that matter had to be represented by a two-index tensor that combined the pressures, currents, and densities of the matter distribution. This matter distribution, if physically meaningful, should also satisfy a continuity equation: $\nabla_{a}T^{ab}=0$, which basically says that matter is neither created nor destroyed in the distribution, and that the time rate of change in a current is the gradient of pressure. When Einstein was writing his field equations down, he wanted some quantity created from the metric tensor that also satisfied this (call it $G^{ab}$) to set equal to $T^{ab}$. But this means that $\nabla_{a}G^{ab} =0$. It turns out that there is only one such combination of terms involving first and second derivatives of the metric tensor: $R_{ab} - \frac{1}{2}Rg_{ab} + \Lambda g_{ab}$, where $\Lambda$ is an arbitrary constant. So, this is what Einstein picked for his field equation.
Now, $R_{ab}$ has the same number of indicies as the stress-energy tensor. So, a hand-wavey way of looking at what $R_{ab}$ means is to say that it tells you the "part of the curvature" that derives from the presence of matter. Where does this leave the remaining components of $R_{abc}{}^{d}$ on which $R_{ab}$ does not depend? Well, the simplest way (not COMPLETELY correct, but simplest) is to call these the parts of the curvature derived from the dynamics of the gravitational field itself--an empty spacetime containing only gravitational radiation, for example, will satisfy $R_{ab}=0$ but will also have $R_{abc}{}^{d}\neq 0$. Same for a spacetime containing only a black hole. These extra components of $R_{abc}{}^{d}$ give you the information about the gravitational dynamics of the spacetime, independent of what matter the spacetime contains.
This is getting long, so I'll leave this at that.
Dear John, let me post the same thing that Marek has said as a standard answer.
Einstein's equations are not equations for the Riemann tensor because the Riemann tensor's components are not independent fields. Instead, Einstein's equations are differential equations for the metric tensor.
In 4 dimensions, the metric tensor has 10 components - a symmetric tensor - and Einstein's equations have 10 components - a symmetric tensor - too. It doesn't matter that the Riemann tensor has 20 components because these 20 functions of space and time are calculated from the 10 component functions of the metric tensor and its (first and second) derivatives.
In fact, the 10-10 counting is oversimplified. Four "differential combinations" of Einstein's equations vanish identically because $\nabla_\mu R^{\mu\nu}=0$ is an identity (that always holds, even if the equations of motion are not satisfied). The same identity holds for the corresponding other tensors that are added to the Ricci tensor in Einstein's equations.
So instead of 10 equations, the Riemann equations are, in some sense, just 6 independent equations. That means that they don't determine the metric completely: they leave 4 functions undetermined and these are exactly the 4 functions that you may choose arbitrarily to specify a diffeomorphism, mapping one solution into another (equivalent) solution.
Up to the coordinate transformations which are always allowed to be made, initial conditions for the metric and its first derivative determine the metric tensor - and therefore the whole Riemann tensor - everywhere in the future. One doesn't need any Newtonian equations as a "mandatory supplement" in general relativity. That doesn't mean that the Newtonian limit is unimportant: of course, it is one of the most important approximate consequences of general relativity.
Best Answer
The Einstein equations are some of the most complicated PDE's people study. There is no shortcut for this, you just have to do all the horrific algebra. Start with the trace-reversed Einstein equation \begin{equation} R_{\mu \nu}=8\pi G(T_{\mu \nu}-\frac{1}{2}Tg_{\mu \nu}) \end{equation} Use the equation for Ricci in terms of the Christoffel Connection \begin{equation} R_{\mu \nu}=2\Gamma^\alpha_{\mu [\nu,\alpha]}+2\Gamma^{\alpha}_{\lambda[\alpha}\Gamma^\lambda_{\nu]\mu} \end{equation} Plug in \begin{equation} \Gamma^i_{kl}=\frac{1}{2}g^{im}(g_{mk,l}+g_{ml,k}-g_{kl,m}) \end{equation} In the end you will just have lots of terms with up to two derivatives of the metric.