Let me show this specifically for the finite element discretization of the Laplace equation:
$$
-\frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial}{\partial r} u(r,\theta)\right)
- \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} u(r,\theta)
= f(r,\theta).
$$
The finite element method is based on the weak formulation, which you obtain by multiplying the equation by a test function $\varphi(r,\theta)$ and integrating over the domain, then integrate by parts. You'd think that that leads to
$$
\int_0^R \int_0^{2\pi}
\varphi(r,\theta)
\left[
-\frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial}{\partial r} u(r,\theta)\right)
- \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} u(r,\theta)
\right]
\, d\theta dr
= \int_0^R \int_0^{2\pi} \varphi(r,\theta) f(r,\theta) \, d\theta dr,
$$
followed by integration by parts.
But you don't have to do it that way -- the above approach simply assumes that $r$ and $\theta$ are independent variables. Instead, you use the proper area element, which is $2\pi r \, dr \, d\theta$ and you get the following form instead:
$$
\int_0^R \int_0^{2\pi}
\varphi(r,\theta)
\left[
-\frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial}{\partial r} u(r,\theta)\right)
- \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} u(r,\theta)
\right]
\, 2\pi r \, d\theta dr
\\
= \int_0^R \int_0^{2\pi} \varphi(r,\theta) f(r,\theta) \, 2\pi r \, d\theta dr.
$$
The $2\pi$ cancel on the two sides, and so what you have left is this:
$$
\int_0^R \int_0^{2\pi}
\varphi(r,\theta)
\left[
-\frac{\partial}{\partial r} \left(r \frac{\partial}{\partial r} u(r,\theta)\right)
- \frac{1}{r} \frac{\partial^2}{\partial \theta^2} u(r,\theta)
\right]
\, d\theta dr
= \int_0^R \int_0^{2\pi} \varphi(r,\theta) f(r,\theta) \, r \, d\theta dr.
$$
After integration by parts, you then have the following (omitting boundary terms for simplicity):
$$
\int_0^R \int_0^{2\pi}
r \frac{\partial \varphi(r,\theta)}{\partial r}
\frac{\partial u(r,\theta)}{\partial r}
+ \frac{1}{r}
\frac{\partial \varphi(r,\theta)}{\partial \theta}
\frac{\partial u(r,\theta)}{\partial \theta}
\, d\theta dr
= \int_0^R \int_0^{2\pi} \varphi(r,\theta) f(r,\theta) \, r \, d\theta dr.
$$
As you can see, at least from in front of the the $r$ derivatives, the singular weight has disappeared. If you choose appropriate quadrature rules, then the quadrature points for the second term will never lie at $r=0$, and so the still singular weight will not matter in the $\theta$-derivative term.
