The cartesian metric tensor is $ \left[ \matrix{ 1 & 0 \cr 0 & 1 } \right] $ while the polar metric tensor is $ \left[ \matrix{ 1 & 0 \cr 0 & r^2 } \right] $. Since both are flat planes, is there any way we can calculate the second one from the first one using a Jacobian variable transformation where $ J=\left[ \matrix{ \cos(\theta) & -r\sin(\theta) \cr \sin(\theta) & r\cos(\theta) } \right] $?
This wikipedia page https://en.wikipedia.org/wiki/Metric_tensor suggests in the section coordinate transformation that we can use the formula $g_{\mu' \nu'}=J^T \cdot g_{\mu \nu} \cdot J$. If this is the solution, where does the formula come from?
The metric tensor (field) $g_{ij}$ is an example of a (0,2)-tensor and hence we can transform the components using the standard transformation law for (0,2)-tensors.
However, since I suspect you are not familiar with tensors I will try to give you a brief introduction to the usual notation of tensors (range convention and summation convention) and give you some feeling for the transformations of tensors.
The polar coordinate transformation $\bar{x}^j=\bar{x}^j(x^h)$ is
$$\bar{x}^1=x^1\cos(x^2)\\\bar{x}^2=x^1\sin(x^2) \tag{1}$$ It might feel more familiar to you if we write it like this $$x=r\cos(\theta)\\y=r\sin(\theta)\tag{2}$$
If we differentiate $(1)$ we get $$d\bar{x}^j=\frac{\partial \bar{x}^j }{\partial x^h}dx^h$$ Where we used the Einstein summation convention; when a small Latin suffix is repeated in a term, summation with respect to that suffix is understood. In our case, this means
$$dx=\frac{\partial x }{\partial r}dr+\frac{\partial x }{\partial \theta}d\theta\\dy=\frac{\partial y }{\partial r}dr+\frac{\partial y }{\partial \theta}d\theta$$
The infinitesimal displacement is the prototype of a class of geometrical objects which are called contravariant tensors.
Notice how we use the Jacobian of the transformation to transform the vector.
In general a set of quantities $T^r$ associated with a point $P$, are said to be the components of a contravariant vector if they transform, on change of coordinates, according to the equation $$\bar{T}^r=\frac{\partial \bar{x}^r}{\partial x^h}T^h$$
It should be well known from the theory of implicit functions,that $\bar{x}^j=\bar{x}^j(x^h)$ may be solved to read $x^h=x^h(\bar{x}^j)$, and in a similar fashion $$dx^h=\frac{\partial x^h}{\partial \bar{x}^j}d\bar{x}^j$$
In much the same way, a set of $n^{2}$ quantities $g_{ij}$ is said to constitute the components of a tensor of type $(0,2)$ at a point $P$, if under the coordinate transformation $\bar{x}^j=\bar{x}^j(x^h)$ these quantities transform according to the law $${g}_{ij}=\frac{\partial \bar{x}^r }{\partial x^i }\frac{\partial \bar{x}^s}{\partial x^j}\bar{g}_{rs}\tag{3}$$
The metric tensor (field) $g_{ij}$ is an example of a $(0,2)$-tensor and hence we can transform it from one coordinate system to another using the standard transformation law $(3)$ for $(0,2)$-tensors.
For the polar coordinate system the Jacobian of the transformation
$$J^j_h=\frac{\partial \bar{x}^j}{\partial x^h}=\begin{bmatrix}\cos(\theta) & -r\sin(\theta)\\ \sin(\theta) & r\cos(\theta) \end{bmatrix}$$
Is a $2x2$-matrix and using the summation convention described above you should be able to figure out that (3) is just $$g=J^T\bar{g}J$$
For example, $g_{22}=g_{\theta \theta}$ is (with $\bar{g}_{11}=\bar{g}_{22}=1\, \bar{g}_{12}=\bar{g}_{21}=0$, that is $\bar{g}$ is the identity matrix in cartesian coordinates):
$$g_{22}=\frac{\partial \bar{x}^1 }{\partial x^2}\frac{\partial \bar{x}^1}{\partial x^2}\bar{g}_{11}+\frac{\partial \bar{x}^2 }{\partial x^2}\frac{\partial\bar{x}^2}{\partial x^2}\bar{g}_{22}=r^2\sin^2(\theta)+r^2\cos^2(\theta)=r^2$$