My discussion essentially follows this article.
Form Factors are an intuitive and simple tool used to describe the scattering particles from extended targets. Here I'm going to show how the Form Factor comes about in the context of the scattering of spinless electrons
As with many scattering experiments, the quantity we are interested in is the
differential cross section $\frac{d\sigma}{d\Omega}$ of our scattered electrons off our target.
The differential cross section is related to the scattering amplitudes through the relation:
$$\frac{d\sigma}{d\Omega}=\frac{k}{k_i}|f(\theta,\phi)|^2$$
where $\theta$ is the scattered angle.
The scattering amplitudes $f(\theta,\phi)$ can be obtained in approximate form using the Born Approximation. To first order (and up to a normalization) the Born Approximation can be written as:
$$f_{B1}=\int\phi_{k_f}^*(\vec{r})V(\vec{r})\phi_{k_i}(\vec{r})d^3\vec{r}$$
In the first Born Approximation the initial incoming wave and the outgoing waves are assumed to be plane waves of the form:
\begin{align}
\phi_{k_i}(\vec{r})=e^{ik_i\cdot r}\\
\phi_{k_f}(\vec{r})=e^{ik_f\cdot r}
\end{align}
We can describe an extended charge distribution by $Ze\rho(\vec{r})$ with
$$\int\rho(\vec{r})d^3\vec{r}=1$$
In this case, the potential experienced by an electron located at $\vec{r}$ is given by the Coloumb potential:
$$V(\vec{r})=-\frac{Ze^2}{4\pi\epsilon_0}\int\frac{\rho(\vec{r}')}{|r-r'|}d^3\vec{r}'$$
And substituting this potential into the general expression for the first Born Approximation to the scattering amplitudes $f(\theta,\phi)$ gives
$$f_{B1}=-\frac{Ze^2}{4\pi\epsilon_0}\int e^{\frac{iq\cdot r}{h}}\int\frac{\rho(\vec{r}')}{|r-r'|}d^3\vec{r}d^3\vec{r}'$$
Making the substitution $\vec{R}=\vec{r}-\vec{r}'$ and noticing that $d^3\vec{R}=d^3\vec{r}$
$$f_{B1}=-\frac{Ze^2}{4\pi\epsilon_0}\int \frac{e^{\frac{iq\cdot \vec{R}}{h}}}{\vec{R}}d^3\vec{R} \left[\int e^{\frac{iq\cdot \vec{r}'}{h}}\rho(\vec{r}')d^3\vec{r}'\right]$$
This bracked factor is known as the Form Factor, $F(q)$.
$$F(q)=\int e^{\frac{iq\cdot \vec{r}'}{h}}\rho(\vec{r}')d^3\vec{r}' $$
It can be shown that when the expression for $f_{B1}$ is used to determine $\frac{d\sigma}{d\Omega}$, that:
$$\frac{d\sigma}{d\Omega}=\left(\frac{Ze}{4E}\right)^2\frac{1}{sin^4(\theta/2)}|F(q)|^2$$
This expression can be interpreted intuitively as Rutherford scattering modulated by the square of the Form Factor. In other words, electron scattering off an extended source is equal to scattering off a point source modulated by the form factor.
That the total cross section is infinite just means that every charged particle that passes by the (bare) nucleus is scattered to some extent. This is a consequence of the Coulomb potential being long range. Classically, it is sufficient for the potential to be nonzero for all radii in order to have an infinite total cross section. Quantum mechanically, if the potential approaches zero rapidly enough, the total cross section will be finite. However, the Coulomb potential goes to zero slowly at infinity, and the resulting Rutherford cross section (which is correct both classically and quantum mechanically) has an infinite total cross section.
Best Answer
Since the Coulomb potential is long ranged, incident states with arbitrarily large impact parameters are scattered by small angles. Taking the impact parameter to infinity, the scattering angle goes to zero, so essentially there are an infinite number of forward scattering events for which the total cross section must account.