Here is how we can recover Newton's laws for gravitation from General Relativity.

We begin with a metric $g_{\mu\nu}$ that is a perturbation $h_{\mu\nu}$ (with $|h_{\mu\nu}|\ll 1$) of the Minkowski-metric $\eta_{\mu\nu}$:

$$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}.$$

Further, we assume that the gravitational field is approximately static, hence time derivatives are zero.

The starting point is Einstein's field equations for gravity:

$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=8\pi GT_{\mu\nu},$$

with the Ricci-tensor given by $R_{\mu\nu}=\partial_\alpha\Gamma^\alpha_{\mu\nu}-\partial_\nu\Gamma^\alpha_{\mu\alpha}+\Gamma^\alpha_{\mu\nu}\Gamma^\beta_{\alpha\beta}-\Gamma^\alpha_{\mu\beta}\Gamma^\beta_{\alpha\nu}$, $\Gamma_{\mu\nu}^\alpha=\frac{1}{2}g^{\alpha\beta}(\partial_\mu g_{\nu\beta}+\partial_\nu g_{\mu\beta}-\partial_\beta g_{\mu\nu})$ are the Christoffel-symbols associated with $g_{\mu\nu}$, and the metric signature is $[+,-,-,-]$.

In the weak field, the Ricci-tensor simplifies to

$$R_{\mu\nu}\simeq\partial_\alpha\Gamma^\alpha_{\mu\nu}-\partial_\nu\Gamma^\alpha_{\mu\alpha}.$$

Moreover,

$$\Gamma_{\mu\nu}^\alpha\simeq \frac{1}{2}\eta^{\alpha\beta}(\partial_\mu h_{\nu\beta}+\partial_\nu h_{\mu\beta}-\partial_\beta h_{\mu\nu}),$$

and

$$R\simeq\eta^{\mu\nu}R_{\mu\nu}.$$

With these approximations, the field equations now read

$$2\partial_\alpha\Gamma^\alpha_{\mu\nu}-2\partial_\nu\Gamma^\alpha_{\mu\alpha}-\eta_{\mu\nu}\eta^{\kappa\lambda}\partial_\alpha\Gamma^\alpha_{\kappa\lambda}+\eta_{\mu\nu}\eta^{\kappa\lambda}\partial_\lambda\Gamma^\alpha_{\kappa\alpha}=16\pi GT_{\mu\nu},$$

or

$$\eta^{\alpha\beta}(\partial_\alpha\partial_\mu h_{\nu\beta}+\partial_\alpha\partial_\nu h_{\mu\beta}-\partial_\alpha\partial_\beta h_{\mu\nu}-\partial_\mu\partial_\nu h_{\alpha\beta}-\eta_{\mu\nu}\eta^{\kappa\lambda}\partial_\alpha\partial_\kappa h_{\lambda\beta}+\eta_{\mu\nu}\eta^{\kappa\lambda}\partial_\alpha\partial_\beta h_{\kappa\lambda})=16\pi GT_{\mu\nu},$$

For simplicity, we introduce

$$\bar{h}_{\mu\nu}=h_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}\eta^{\alpha\beta}h_{\alpha\beta}.$$

Then we get

$$

\eta^{\alpha\beta}\partial_\alpha\partial_\mu\bar{h}_{\nu\beta}

+\eta^{\alpha\beta}\partial_\alpha\partial_\nu\bar{h}_{\mu\beta}

-\eta^{\alpha\beta}\partial_\alpha\partial_\beta\bar{h}_{\mu\nu}

-\eta^{\alpha\beta}\eta_{\mu\nu}\eta^{\kappa\lambda}\partial_\alpha\partial_\kappa\bar{h}_{\lambda\beta}

=16\pi GT_{\mu\nu}.$$

We know that without loss of generality, we can impose the Lorenz-gauge:

$$\eta^{\alpha\beta}\partial_\alpha\bar{h}_{\nu\beta}=0.$$

When we do so, the field equations become just

$$-\eta^{\alpha\beta}\partial_\alpha\partial_\beta\bar{h}_{\mu\nu}=16\pi GT_{\mu\nu}.$$

When the metric is static ($\partial_0\bar{h}_{\mu\nu}=0$), we get

$$\nabla^2\bar{h}_{\mu\nu}=16\pi GT_{\mu\nu},$$

where $\nabla$ is the three-dimensional vector gradient operator.

Now let $h_{\mu\nu}=\mathrm{diag}[2\phi,2\phi,2\phi,2\phi]$. In this case, $\bar{h}_{\mu\nu}=h_{\mu\nu}+2\phi\eta_{\mu\nu}=\mathrm{diag}[4\phi,0,0,0]$, and recognizing that $T_{00}=\rho$ is just the matter density, the field equations turn into Poisson's equation for gravity:

$$\nabla^2\phi=4\pi G\rho.$$

All other components of $T_{\mu\nu}$ are zero. The equations $T_{0i}=0$, $T_{ij}=0~(i,j=1..3)$ state that in this approximation, insofar as gravity is concerned, momenta, pressure and stresses are negligible.

We can also derive the Newtonian acceleration law directly. We begin with the geodesic equation of motion, itself a direct consequence of Einstein's field equations:

$$\frac{d^2x^\alpha}{d\tau^2}+\Gamma_{\mu\nu}^\alpha\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0,$$

where $x^\alpha$ is the 4-vector describing a test particle and $\tau$ is proper time. When the motion is slow, we can neglect $dx^i/d\tau$ ($i=1..3$) compared to $dt/d\tau$ ($t=x^0$):

$$\frac{d^2x^\alpha}{d\tau^2}+\Gamma_{00}^\alpha\left(\frac{dt}{d\tau}\right)^2=0.$$

In a stationary gravitational field, all time derivatives vanish, thus $\Gamma^\mu_{00}=-\frac{1}{2}g^{\mu\nu}\partial g_{00}/\partial x^\nu$. Once again writing the metric in the form $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$, we have, approximately,

$$\Gamma^\mu_{00}=-\frac{1}{2}\eta^{\mu\nu}\frac{\partial h_{00}}{\partial x^\nu}.$$

The temporal term in this equation amounts to $dt/d\tau$ being constant in time, which is as it should be, given the static metric. Let $h_{\mu\nu}=\mathrm{diag}(2\phi,2\phi,2\phi,2\phi)$. Then the rest of the geodesic equation reads

$$\frac{d^2\vec{x}}{dt^2}=-\nabla\phi,$$

which is just the Newtonian gravitational acceleration law in a potential given by $\phi$.