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
When going around a "small" path on a curved surface, the angle you come up short is given by the local Guass-Bonnet theorem. Usually the G-B theorem comes up in connection with topology, involving integrals over the whole manifold. But that is just one facet of it. The local version I find useful at the moment is:
$\iint_{enclosed} K dA = 2\pi - \sum \gamma_i $
Curvature $K$ is 1/(radius of curvature) per dimension, and we multiply together two dimensions. For a sphere, it's constant so the left hand side is just $KA$, $A$ being the area enclosed by your closed-path travel. $\gamma_i$ is the angle of turning at each point $i$ where you make a turn. (For a continuously curving path, there'd be an integral.) Going straight through a point $i$ is $\gamma_i=0$.
If you go around a square of side length L, with L much smaller than R, the area is $L^2$ (pretty close to it.) You think you'd have four turns of 90 degrees, adding to 360 (which is $2\pi radians), but Gauss-Bonnet says no. You'll make three 90 degree turns, and one that's not quite, if you travel a closed path ending at your starting point. Alternatively, you can make strict 90 degree turns four times, going exactly distance L for each side, and then measure how far away you end up from the starting point, and do some extra math. G-B still applies, just so not directly.
At this point, if you have a notion of how small an angular error you can measure, you can plug in numbers. $R_{Earth}$ is about 4000 miles (6400 km). Area $A$ is $L^2$ and $K=1/R^2$. On Earth, if we do this experiment on a reasonably human scale, we hope a few hundred meters or a few kilometers, we expect close to the same value for $A$. Say you can measure, or steer, to about half a degree accuracy. That's 1/100th radian, roughly. We already intend to make four ideal 90 degree turns (the $2\pi$ in the G-B formula) but expect to measure some small angular deficit
$\delta = 2\pi-\sum \gamma$.
when we finish the trip at the start. To get an idea what size trip we must make, just set $\delta = 0.04$ for four steering measurements good to .01 radians. (A naive estimate, not a proper error analysis.) Then,
$ {{L^2}\over{R^2}} = 0.04$
I'll let you grab a calculator and punch in the numbers, and try more accurate angle.
Note this is a rough "back-of-the-envolope" calculation. Better thought-out math would be needed for a real experiment.