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.
Your example is basically the motion of test particles constrained to a 2D surface (in this case, the surface of the Earth) in flat spacetime. If the surface in question is sufficiently smooth, then we can define a spatial metric $\gamma$ which is inherited from the embedding of the surface in flat 3D space.
Here's how this works for the 2-sphere. We can coordinatize points on the 2-sphere using the angular coordinates $(\theta,\phi)$ such that
$$\pmatrix{x(\theta,\phi)\\y(\theta,\phi)\\z(\theta,\phi)} = \pmatrix{R\sin(\theta)\cos(\phi)\\ R\sin(\theta)\sin(\phi) \\ R \cos(\theta)}$$
The 2-sphere is defined by the equation $x^2+y^2+z^2 = R^2$. The line element in flat 3D space is given by $\mathrm d\sigma^2=\mathrm dx^2+\mathrm dy^2+\mathrm dz^2$. Plugging in the coordinate expressions above and doing some algebra yields
$$\mathrm d\sigma^2 = R^2 \mathrm d\theta^2 + R^2\sin^2(\theta) \mathrm d\phi^2$$
which is the induced line element on the 2-sphere. The components of the corresponding induced metric $\gamma$ can be read off as
$$\gamma_{ij} = \pmatrix{R^2 & 0 \\ 0 & R^2\sin^2(\theta)}$$
The full spacetime line element and metric in the coordinates $(t,\theta,\phi)$ are then simply
$$\mathrm ds^2 = -c^2\mathrm dt^2 + R^2 \mathrm d\theta^2 + R^2\sin^2(\theta) \mathrm d\phi^2$$
$$g_{ij} = \pmatrix{-c^2 & 0 & 0 \\ 0 & R^2 & 0 \\ 0& 0 & R^2\sin^2(\theta)}$$
and the inverse metric components are
$$g^{ij} = \pmatrix{-\frac{1}{c^2} & 0 &0 \\ 0 & \frac{1}{R^2} & 0 \\ 0 & 0 & \frac{1}{R^2\sin^2(\theta)}}$$
From here we can compute the Christoffel symbols, which is a straightforward exercise (the only non-constant component of the metric tensor is $g_{\phi\phi}$, so almost all of them vanish). That's all we need for the geodesic equation, so if we want to understand the motion of test particles then we're basically done.
What are the indispensable insights gained from each of the following steps for this particular example: Metric Tensor -> Christoffel Symbols -> Riemann Curvature -> Ricci Curvature in order to make useful predictions in this case?
Once we have the Christoffel symbols, we don't need the Riemann or Ricci curvatures to predict how particles will move. As in the example above, if you are handed the metric (e.g. if you are considering constrained motion in flat spacetime) then by taking a few derivatives you can compute the $\Gamma$'s, and therefore have the equations of motion for test particles.
The point you may be missing is that in GR, we generally don't have the metric; instead, we start with the stress-energy tensor $T_{\mu\nu}$ which tells us the distribution of energy and momentum in the spacetime we are considering. From there, note that the $\Gamma$'s involve the first derivatives of the (unknown!) metric, while the Riemann tensor involves its second derivatives. The Ricci and Einstein tensors are just algebraic combinations of the above, so the expression
$$G_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$
is an elaborate system of second-order, nonlinear, coupled PDE's for the components $g_{ij}$ of the unknown metric.
The typical way to solve these equations is to use symmetry arguments to constrain the form of the unknown metric. For example, assuming a static and spherically symmetric vacuum spacetime leads to coordinates $(t,r,\theta,\phi)$ in which the line element takes the form
$$\mathrm ds^2 = -A(r)\mathrm dt^2 + B(r)\mathrm dr^2 + r^2(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2)$$
So the problem of finding the unknown metric has already been reduced to finding two unknown functions of the single coordinate $r$. This is vastly simpler than the general case, in which you have 10 independent components $g_{ij}$, each of which is a function of all 4 coordinates. Plugging this in to Einstein's equations and turning the crank is still a tedious exercise (you can find it worked out in full detail here), but it ultimately yields very simple equations for $A$ and $B$:
$$A(r) B(r) = c^2$$
$$\frac{d}{dr}\left(r^2 A'(r)\right) = 0$$
From there, one can demand that this reproduce Newtonian gravity in the large-$r$ limit; the result is the Swarzschild metric. Once you have that, you can compute the Christoffel symbols and obtain the equations of motion for test particles.
So in summary, the entire point of solving Einstein's equations - at least in terms of predicting the motion of test particles - is to obtain the components of the metric, and thereby the Christoffel symbols which appear in the particles' equations of motion. If you already have the metric/$\Gamma$'s, the Riemann/Einstein tensors don't give you any additional insight into how particles will move.
Best Answer
The Riemann curvature tensor is defined by:
$$R_{abc}{}^{d}\omega_{d} = \nabla_{[a}\nabla_{b]}\omega_{c}$$
for arbitrary $\omega_{a}$. If you work this out in terms of Chistoffel symbols (or through a more sophisticated proof), you'll discover that it has the following symmetries:
$$R_{abcd} = -R_{bacd} = -R_{abdc} = R_{cdab}$$
From this, you can work out that, in two dimensions, there is only one independent component of the Riemann tensor. This component is directly proportional to the Gaussian curvature of the two-dimensional surface. In higher dimensions, the Riemann curvature extends the Gaussian curvature, but you need all of these extra components because:
In two dimensions, all of this is taken care of for you, because there is only one plane, but in the general case, you need all four components for anything to make consistent sense.