It is useful practice of the techniques developed in the previous chapters to derive (8.3) and (8.4). We outline the various steps below. The electromagnetic field due to $Z e \rho(\mathbf{x})$ is $A^\mu=(\phi, 0)$ where, using (6.59),
$$
\nabla^2 \phi=-Z e \rho(\mathbf{x})
$$
Use (6.4) and (6.6) to show that the scattering amplitude is (see also Section 7.1)
$$
T_{f i}=-i 2 \pi \delta\left(E_f-E_i\right)\left(-e \bar{u}_f \gamma_0 u_i\right) \int e^{i \mathbf{q} \cdot \mathbf{x}} \phi(\mathbf{x}) d^3 x
$$
Justify
$$
\int e^{i \mathbf{q} \cdot \mathrm{x}} \nabla^2 \phi d^3 x=-|\mathbf{q}|^2 \int e^{i \mathbf{q} \cdot x} \phi d^3 x
$$
and hence show that the integral in (8.5) is $Z e F(\mathbf{q}) /|\mathbf{q}|^2$, see (7.9). Following the arguments of Section 4.3 , verify that the differential cross section from a fixed target is
$$
d \sigma=\frac{\left|T_{f i}\right|^2}{T} \frac{d^3 k_f}{(2 \pi)^3 2 E_f}\left(\frac{1}{v 2 E_i}\right),
$$
with
$$
d^3 k_f \delta\left(E_f-E_i\right)=k E d \Omega .
$$
Summing final, and averaging initial, electron spins give
$$
\frac{1}{2} \sum_{s_f, s_i}\left|\bar{u}_f \gamma_0 u_i\right|^2=4 E^2\left(1-v^2 \sin ^2 \frac{\theta}{2}\right),
$$
where $\theta$ is the angle introduced in Section 7.1. Check this answer with (6.25). Putting all this together yields the advertised result
$$
\frac{d \sigma}{d \Omega}=\left(\frac{d \sigma}{d \Omega}\right)_{\text {Mott }}|F(\mathbf{q})|^2,
$$
with the form factor given by (8.3).