The question is extremely interesting, and to be properly answered it would require notions of Riemannian geometry (that I may add later) that probably are a bit too sophisticated. In any case, one can do the following for the 2 dimensional case (the 3d can be found here https://planetmath.org/DerivationOfTheLaplacianFromRectangularToSphericalCoordinates):
Recall that Laplace's equation in $\mathbb{R}^{2}$ in terms of the usual (i.e., Cartesian) $(x, y)$ coordinate system is:
$$
\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=u_{x x}+u_{y y}=0
$$
The Cartesian coordinates can be represented by the polar coordinates as follows:
$$
\left\{\begin{array}{l}
x=r \cos \theta \\
y=r \sin \theta
\end{array}\right.
$$
Let us first compute the partial derivatives of $x, y$ w.r.t. $r, \theta$ :
$$
\left\{\begin{array}{ll}
\frac{\partial x}{\partial r}=\cos \theta, & \frac{\partial x}{\partial \theta}=-r \sin \theta \\
\frac{\partial y}{\partial r}=\sin \theta, & \frac{\partial y}{\partial \theta}=r \cos \theta
\end{array}\right.
$$
To do so, let's compute $\frac{\partial u}{\partial r}$ first. We will use the Chain Rule since $(x, y)$ are functions of $(r, \theta)$ as shown above.
\begin{aligned}
\frac{\partial u}{\partial r} &=\frac{\partial u}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial u}{\partial y} \frac{\partial y}{\partial r} \\
&=\frac{\partial u}{\partial x} \cos \theta+\frac{\partial u}{\partial y} \sin \theta \quad \text { using } \\
&=\cos \theta \frac{\partial u}{\partial x}+\sin \theta \frac{\partial u}{\partial y}
\end{aligned}
Now, let's compute $\frac{\partial^{2} u}{\partial r^{2}}$. Noticing that both $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ are functions of $(x, y)$
we have
$$
\begin{aligned}
\frac{\partial^{2} u}{\partial r^{2}} &=\cos \theta \frac{\partial}{\partial r} \frac{\partial u}{\partial x}+\sin \theta \frac{\partial}{\partial r} \frac{\partial u}{\partial y} \\
&=\cos \theta\left(\frac{\partial}{\partial x} \frac{\partial u}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial}{\partial y} \frac{\partial u}{\partial x} \frac{\partial y}{\partial r}\right)+\sin \theta\left(\frac{\partial}{\partial x} \frac{\partial u}{\partial y} \frac{\partial x}{\partial r}+\frac{\partial}{\partial y} \frac{\partial u}{\partial y} \frac{\partial y}{\partial r}\right) \\
&=\cos ^{2} \theta \frac{\partial^{2} u}{\partial x^{2}}+2 \cos \theta \sin \theta \frac{\partial^{2} u}{\partial x \partial y}+\sin ^{2} \theta \frac{\partial^{2} u}{\partial y^{2}}
\end{aligned}
$$
Similarly, let's compute $\frac{\partial u}{\partial \theta}$ and $\frac{\partial^{2} u}{\partial \theta^{2}}$.
$$
\begin{aligned}
\frac{\partial u}{\partial \theta} &=\frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta} \\
&=\frac{\partial u}{\partial x}(-r \sin \theta)+\frac{\partial u}{\partial y}(r \cos \theta) \\
&=-r \sin \theta \frac{\partial u}{\partial x}+r \cos \theta \frac{\partial u}{\partial y}
\end{aligned}
$$
\begin{aligned}
\frac{\partial^{2} u}{\partial \theta^{2}}=&-r \cos \theta \frac{\partial u}{\partial x}-r \sin \theta \frac{\partial}{\partial \theta} \frac{\partial u}{\partial x}-r \sin \theta \frac{\partial u}{\partial y}+r \cos \theta \frac{\partial}{\partial \theta} \frac{\partial u}{\partial y} \\
=&-r \cos \theta \frac{\partial u}{\partial x}-r \sin \theta\left(\frac{\partial}{\partial x} \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta}+\frac{\partial}{\partial y} \frac{\partial u}{\partial x} \frac{\partial y}{\partial \theta}\right)-r \sin \theta \frac{\partial u}{\partial y}+r \cos \theta\left(\frac{\partial}{\partial x} \frac{\partial u}{\partial y} \frac{\partial x}{\partial \theta}+\frac{\partial}{\partial y} \frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta}\right) \\
=&-r \cos \theta \frac{\partial u}{\partial x}-r \sin \theta\left(\frac{\partial^{2} u}{\partial x^{2}}(-r \sin \theta)+\frac{\partial^{2} u}{\partial x \partial y} r \cos \theta\right) \\
&-r \sin \theta \frac{\partial u}{\partial y}+r \cos \theta\left(\frac{\partial^{2} u}{\partial x \partial y}(-r \sin \theta)+\frac{\partial^{2} u}{\partial y^{2}} r \cos \theta\right) \\
=&-r\left(\cos \theta \frac{\partial u}{\partial x}+\sin \theta \frac{\partial u}{\partial y}\right)+r^{2}\left(\sin ^{2} \theta \frac{\partial^{2} u}{\partial x^{2}}-2 \cos \theta \sin \theta \frac{\partial^{2} u}{\partial x \partial y}+\cos ^{2} \theta \frac{\partial^{2} u}{\partial y^{2}}\right)
\end{aligned}
Dividing both sides by $r^{2}$ and using the equation above, we have
$$
\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}=-\frac{1}{r} \frac{\partial u}{\partial r}+\sin ^{2} \theta \frac{\partial^{2} u}{\partial x^{2}}-2 \cos \theta \sin \theta \frac{\partial^{2} u}{\partial x \partial y}+\cos ^{2} \theta \frac{\partial^{2} u}{\partial y^{2}}
$$
Finally, adding the last two, using the obvious relation $\cos ^{2} \theta+\sin ^{2} \theta=1$, we have
$$
\frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}=-\frac{1}{r} \frac{\partial u}{\partial r}+\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}
$$
which can be cleaned up as:
$$
\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=\frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}
$$
Hence, Laplace's equation becomes:
$$
u_{x x}+u_{y y}=u_{r r}+\frac{1}{r} u_{r}+\frac{1}{r^{2}} u_{\theta \theta}=0
$$
