Let's work with a 2 dimensional function $f(x, y)$ and lets assume its continuously differentiable infinitely many times and we don't any problems from that direction.
the partial derivative of $f$ with respect to $x$ at point $(a,b)$ is defined as $\frac{\partial f}{\partial x}(a,b) = \lim_{h \to 0}\frac{f(a+h, b) - f(a,b)}{h}$.
That's fine, I get that, for higher order as well.
But now suppose $(u, v)$ is the rotation of $(x,y)$ by $\theta$ degrees counter clockwise.
One of the properties of Laplacian operator is that it is rotation invariant. Meaning $\frac{\partial ^2 f}{\partial x^2}(a,b) + \frac{\partial ^2 f}{\partial y^2}(a,b) = \frac{\partial ^2 f}{\partial u^2}(a,b) + \frac{\partial ^2 f}{\partial v^2}(a,b)$. I even proved it myself with chain rule, it checks out.
But what the heck does $\frac{\partial f}{\partial u}$ even mean? With respect to which coordinate is it a partial derivative? I feel like I'm missing the bigger picture.
The $f$ in your question really stands for two different functions. Let’s separate them. Letting $\phi:(u,v)\mapsto(x,y)$ be the rotation, the second function that you’re dealing with is really $g = f\circ\phi$. The partial derivative ${\partial g\over\partial u}$ should then make perfect sense in terms of coordinates, and can be computed via the chain rule. That’s what the somewhat imprecise notation ${\partial f\over\partial u}$ means here.
If you’re familiar with directional derivatives, here’s another way of looking at it. Partial derivatives are just special directional derivatives taken in the directions of the coordinate axes. From that point of view, you can think of ${\partial f\over\partial u}$ as equivalent to $D_uf$, the directional derivative of $f$ in the $u$ direction.