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.

For this derivation, following standard textbooks, we set $h_{\mu\nu} = {\rm diag}[2\phi,2\phi,2\phi,2\phi]$ but this was an ad hoc choice. Assuming approximate spatial homogeneity and isotropy, perhaps we should have used instead $h_{\mu\nu} = {\rm diag}[2\phi,2\psi,2\psi,2\psi]$. In this case, we'd have

\begin{align}
\bar{h}_{\mu\nu}=h_{\mu\nu}+(3\psi-\phi)\eta_{\mu\nu}={\rm diag}[\phi+3\psi,\phi-\psi,\phi-\psi,\phi-\psi].
\end{align}

Using $\varphi=\phi-\psi$ yields

\begin{align}
\bar{h}_{\mu\nu}=h_{\mu\nu} + (2\phi-3\varphi)\eta_{\mu\nu} = {\rm diag}[4\phi-3\varphi, \varphi, \varphi, \varphi].
\end{align}

But this yields, recognizing that $T_{00}=\rho$ but also $T_{ii}=p$ is the pressure,
\begin{align}
\nabla^2(4\phi-3\varphi)&{}=16\pi G\rho,\\
\nabla^2\varphi&{}=16\pi G p,
\end{align}
or
\begin{align}
\nabla^2\phi = 4\pi G(\rho+3p).
\end{align}
Of course when matter is non-relativistic, $p\ll\rho$ and we get back the Newtonian form of Poisson's equation. But this is not the case, when we deal with radiation $(p=\tfrac{1}{3}\rho)$ or dark energy $p=-\rho$.

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 $\phi$.