Suppose $(M,g_{\mu\nu})$ is some spacetime and $p$ is a point in $M$. We can always choose coordinates so that at p, $g_{\mu\nu} = \eta_{\mu\nu}$ and $\partial_\lambda g_{\mu\nu} = 0$, where $\eta_{\mu\nu}$ is the ordinary flat metric. (Note that this only holds at $p$, not necessarily in a neighborhood). These are called exponential coordinates. In this sense, there is absolutely no meaning to the components of the metric tensor, since at any point we can choose coordinates so that $g_{\mu\nu} = \eta_{\mu\nu}$. In fact, it can be shown that in these coordinates the metric takes the form $$ g_{\mu\nu} = \eta_{\mu\nu} + O(x^2)$$ where the precise coefficients of the $O(x^2)$ terms have to do with the Riemann curvature tensor. Thus the only thing that has any physical meaning is the Riemann curvature tensor, since it is the most basic quantity that can be built out of the metric that cannot be eliminated by a simple change of coordinates.
The interpretation that you seem to be looking for comes from thinking of the metric as being a small perturbation a background metric, i.e. 298
$$ g_{\mu\nu} = \eta_{\mu\nu} + \epsilon h_{\mu\nu} $$ for some sufficiently small $\epsilon$. In this case, we can approximate the full <PERSON> equations by equations linear in $h_{\mu\nu}$, and the equations thus obtained are formally analogous to <PERSON>'s equations for the electromagnetic field (this might be called gravitomagnetodynamics). Furthermore, the geodesic equation reduces to something analogous to the Lorentz force law. In fact, it looks something like "free particle + effective potential", where the effective potential is a certain function of $h_{\mu\nu}$ and its derivatives.
Perhaps a better (and certainly more invariant) way of interpreting the metric is to ask what happens to nearby free particles. That is, if I am in some inertial frame, and I throw a few specks of dust in the air in front of me, what do the trajectories of the particles look like, relative to me? The answer is given by the geodesic deviation equation. Again, it is expressed explicitly in terms of the Riemann curvature tensor. In my opinion (other posters: feel free to disagree), the geodesic deviation equation is the best way of understanding the physical interpretation of the metric tensor, since it gives an explicit relation between the relative acceleration of nearby free particles and components of the curvature tensor. 298