How to shorten the derivation of the Laplacian in polar coordinates?

If you are familiar with the language of differential forms, the Laplacian is $$\star d \star d,$$ where $\star$ is the Hodge star operator and $d$ is exterior derivative.

In polar coordinates, the volume form is $r dr \wedge d\theta$.

The Hodge star operator sends the volume form to $1$ ($\star r dr \wedge d\theta = 1$) and acts on the $1$-form basis as follows:

$$\star dr = r d\theta,\\ \star r d\theta = -dr.$$

So for a function $f$ we have: \begin{align} \star d \star d f &=\star d \star (f_r dr + \frac{f_\theta}{r} rd\theta)\\ &=\star d (f_r r d\theta-\frac{f_\theta}{r} dr)\\ &=\star (f_{rr}r dr \wedge d\theta + f_r dr \wedge d\theta - \frac{f_{\theta\theta}}{r} d\theta \wedge dr)\\ &=\star (f_{rr} + \frac{f_r}{r} + \frac{f_{\theta\theta}}{r^2}) r dr \wedge d\theta\\ &=f_{rr} + \frac{f_r}{r} + \frac{f_{\theta\theta}}{r^2} \end{align}