The geodesic equation is given by,
\begin{equation}
\frac{\mathrm d^2}{\mathrm d\tau^2}x^{\mu}+\Gamma^{\mu}_{\lambda\sigma}\frac{\mathrm dx^{\lambda}}{\mathrm d\tau}\frac{\mathrm dx^{\sigma}}{\mathrm d\tau}=0
\end{equation}
which is a set of 4 equations for $x^{i}$.
$\Gamma^{i}_{jk}$ tells us about the curvature of the space time which can be written in terms of derivatives of the metric:
\begin{equation}
\Gamma^{\mu}_{\alpha\beta}=\frac{1}{2}g^{\mu\nu}(\partial_{\alpha}g_{\gamma\beta}+\partial_{\beta}g_{\gamma\alpha}-\partial_{\gamma}g_{\alpha\beta})
\end{equation}
The question is can we get the standard gravitational law from this, For this let's take the following conditions:
$c=1$ and $v^i\ll 1$
which implies that $\mathrm dx^i\ll \mathrm dt$.
Now the expansion of the second term in the geodesic equation yields,
$$
\Gamma^{\mu}_{\lambda\sigma}\frac{\mathrm dx^{\lambda}}{\mathrm d\tau}\frac{\mathrm dx^{\sigma}}{\mathrm d\tau}~\approx~ \Gamma^{\mu}_{00}\left(\frac{\mathrm dt}{\mathrm d\tau}\right)^2 + 0+\ldots$$
as $\mathrm dx\ll \mathrm dt$.
and also $\Gamma^{\mu}_{00}=\frac{1}{2}g^{\mu\sigma}(\partial_0g_{00}+\partial_0 g_{\sigma 0}-\partial_{\sigma}g_{00})$.
Now $\partial_0$ acting on everything will be 0, thus giving;
$$
\frac{\mathrm d^2}{\mathrm d\tau^2}x^i~=~\frac{1}{2}g^{i\sigma}\partial_{\sigma}g_{00}\left(\frac{\mathrm dt}{\mathrm d\tau}\right)^2\;.$$
Now we can take, $g_{00}=\eta_{00}+h_{00}=-1-2\Phi$, where $\Phi$ is the usual gravitational potential. After this we have to put in some approximations regarding the potential strength and the usual Newtonian gravity appears.
Coming to the other part of the question as to how the Riemann tensor depends on the metric, lets look at the following.
we have the following relation:
\begin{equation}
R^{\sigma}_{\lambda\nu\mu}=\partial_{\mu}\Gamma^{\sigma}_{\lambda\nu}-\partial_{\nu}\Gamma^{\sigma}_{\lambda\mu}+\Gamma^{\sigma}_{\mu\rho}\Gamma^{\rho}_{\lambda\nu}-\Gamma^{\sigma}_{\nu\rho}\Gamma^{\rho}_{\lambda\mu}.
\end{equation}
Now we go to a frame of reference in which the christoffel connection vanishes, i.e., the first derivative of the metric is 0, in which case the Riemann tensor becomes:
\begin{equation}
R^{\sigma}_{\lambda\nu\mu}=g^{\sigma\rho}(\partial_{\mu}\partial_{\lambda}g_{\rho\nu}-\partial_{\mu}\partial_{\rho}g_{\nu\lambda}-\partial_{\nu}\partial_{\lambda}g_{\rho\mu}+\partial_{\nu}\partial_{\rho}g_{\mu\lambda})
\end{equation}
This gives the relation between the metric and the Riemann tensor.
