Geometric interpretation of equating a function of $(x,y)$ to a function of $(r,\theta)$ If $u=f(x,y)$, $x=r\cos(\theta)$,$y=r\sin(\theta)$, then we know that $(\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2=(\frac{\partial u}{\partial r})^2+\frac{1}{r^2}(\frac{\partial u}{\partial {\theta}})^2$. Does this equality have a kind of geometric interpretation?
 A: Yes, and it relates to differential geometry and the fact that manifolds (in this case, $\mathbb{R}^2$) can be reparametrized, or described in different coordinates (in this case, Cartesian vs polar). 
We define a change of variables map $\Phi$ (from polar variables to Cartesian) by $\Phi(r, \theta) = (r\cos\theta, r\sin\theta)$. You can think of $\Phi$ as parametrizing the plane (but formally, we are missing some points in this parametrization. What are they?). In terms of the Cartesian basis vectors, $e_x =  (1, 0)$ and $(e_y) = (0, 1)$ the basis vectors of the polar parametrization are 
\begin{align*}
e_r &= \frac{\partial \Phi}{\partial r} = \cos(\theta)e_x + \sin(\theta)e_y \\
e_\theta &= \frac{\partial \Phi}{\partial \theta} = -r\sin(\theta)e_x + r\cos(\theta)e_y.
\end{align*}
Since we'll use the other direction too, note that then 
\begin{align*}
e_x &= \cos(\theta)e_r -  \frac{1}{r}\sin(\theta)e_\theta \\
e_y &= \sin(\theta)e_r +  \frac{1}{r}\cos(\theta)e_\theta.
\end{align*}
Notation: $u_x = \frac{\partial u}{\partial x}$, etc. Now, the idea is that the vector $f_x e_x + f_y e_y$ is equal to the vector $u_r e_r + \frac{1}{r^2}u_\theta e_\theta$. This is  the underlying geometric reason as to the formula you've written, and if there is disbelieve as to why these are equal, use the chain rule on $u_r,  u_\theta$ and the above formulas for the different basis vectors to verify this.  
Now, in Cartesian coordinates,  the (square of the) length of a vector $v$ is $ds^2(v) = v_x^2  + v_y^2$; this can be written as the familiar $ds^2 = dx^2  + dy^2$. But in polar coordinates, the metric $ds^2$ changes as follows:
$$
ds^2_P = dr^2 +  r^2d\theta^2.
$$
What this means is that to calculate the length squared of a vector $v = v_r e_r  + v_\theta e_\theta$ (so, we are given the vector in polar coordinates!), we need to take $v_r^2$, and add this to $r^2 v_\theta^2$ (not just to $v_\theta^2$, like in Cartesian!). When we do this to $v = u_re_r + \frac{1}{r^2} u_\theta d\theta$, we get 
$$
ds^2_P(v) = u_r^2 + r^2\frac{1}{r^4} u_\theta^2 = u_r^2 + r^{-2}u_\theta^2.
$$
But the length squared is a measurement, which doesn't  depend on the coordinates used to measure it; therefore this equals the length of the vector $u_r e_r + \frac{1}{r^2} u_\theta e_\theta$ as given in Cartesian coordinates, which was $f_x^2 + f_y^2$.
TL;DR: The factor of $r^{-2}$ is intimately related to Riemannian metrics, which is related to  the study of how to calculate lengths of vectors under a change of coordinates.* The formula for this length, in combination with the change of basis formulas to represent the same vector with respect to different bases, exactly gives the formula you've written.
