I don't think you can drive the line element with the jacobian $J$
The Kruskal-Szekeres line element
Beginning with the Schwarzschild line element:
\begin{align*}
&\boxed{ds^2 =\left(1-\frac{r_s}{r}\right)\,dt^2-\left(1-\frac{r_s}{r}\right)^{-1}\,dr^2-r^2\,d\Omega^2}\\\\
r_s &:=\frac{2\,G\,M}{c^2} \,,\quad
\text{for 2 dimension space}\\
ds^2 & =\left(1-\frac{r_s}{r}\right)\,dt^2-\left(1-\frac{r_s}{r}\right)^{-1}\,dr^2
\end{align*}
Step I)
\begin{align*}
&\text{for} \quad ds^2=0\\
0&=\left(1-\frac{r_s}{r}\right)\,dt^2-\left(1-\frac{r_s}{r}\right)^{-1}\,dr^2\,,\Rightarrow\\
\left(\frac{dt}{dr}\right)^2&=\left(1-\frac{r_s}{r}\right)^{-2}\,,\Rightarrow
\quad t(r)=\pm\underbrace{\left[r+r_s\ln\left(\frac{r}{r_s}-1\right)\right]}_{r^*}\\
&\Rightarrow\\
\frac{dr^*}{dr}&=\left(1-\frac{r_s}{r}\right)^{-1}\,,\quad
\frac{dr}{dr^*}=\left(1-\frac{r_s}{r}\right)\,,&(1)
\end{align*}
Step II)
\begin{align*}
&\text{New coordinates}\\
u & =t+r^* \\
v & =t-r^*\\
&\Rightarrow\\
t&=\frac{1}{2}(u+v)\,,\quad dt=\frac{1}{2}(du+dv)\\
r^*&=\frac{1}{2}(u-v)\,,\quad dr^*=\frac{1}{2}(du-dv)\\
dr&=\left(1-\frac{r_s}{r}\right)\,dr^*=\frac{1}{2}\,\left(1-\frac{r_s}{r}\right)
(du-dv) \quad\quad(\text{With equation (1)})\\
\Rightarrow
\end{align*}
\begin{align*}
ds^2 &=\left(1-\frac{r_s}{r}\right)\,du\,dv
\end{align*}
Step III)
\begin{align*}
r^* & =\left[r+r_s\ln\left(\frac{r}{r_s}-1\right)\right]= \frac{1}{2}(u-v)\,\Rightarrow\\
\left(\frac{r}{r_s}-1\right)&=\exp\left(-\frac{r}{r_s}\right)
\,\exp\left(\frac{1}{2\,r_s}(u-v)\right)\\
\left(1-\frac{r_s}{r}\right)&=\frac{r_s}{r}\left(\frac{r}{r_s}-1\right)\\
\,\Rightarrow\\\\
ds^2&=\frac{r_s}{r}\,\exp\left(-\frac{r}{r_s}\right)
\,\exp\left(\frac{1}{2\,r_s}(u-v)\right)\,du\,dv
\end{align*}
Step IV)
\begin{align*}
&\text{New coordinates}\\
U= & -\exp\left(\frac{u}{2\,r_s}\right)
\,,\quad
\frac{dU}{du}=-\frac{1}{2\,r_s}\,\exp\left(\frac{u}{2\,r_s}\right)\\
V= & \exp\left(-\frac{v}{2\,r_s}\right)
\,,\quad
\frac{dV}{dv}=-\frac{1}{2\,r_s}\,\exp\left(-\frac{v}{2\,r_s}\right)\\
\,\Rightarrow\\\\
ds^2&=\frac{4\,r_s^3}{r}\exp\left(-\frac{r}{r_s}\right)
\,dU\,dV
\end{align*}
Step V)
\begin{align*}
&\text{New coordinates}\\
U & =T-X\,,\quad dU=dT-dX \\
V & =T+X\,,\quad dV=dT+dX\\
\,\Rightarrow\\\\
&\boxed{ds^2=\frac{4\,r_s^3}{r}\exp\left(-\frac{r}{r_s}\right)
\left(dT^2-dX^2\right)}
\end{align*}
With Matrices and Vectors
The Kruskal-Szekeres line element
Beginning with :
\begin{align*}
ds^2 & =a\,du\,dv\\
&\Rightarrow\\
g&=\frac{1}{2}\begin{bmatrix}
0 & a \\
a & 0 \\
\end{bmatrix}\\\\
q'&=\begin{bmatrix}
du \\
dv \\
\end{bmatrix}\,,\quad
q=\begin{bmatrix}
u \\
v \\
\end{bmatrix} \,,\quad a=\left(1-\frac{r_s}{r}\right)
\end{align*}
Step I)
\begin{align*}
R&=
\begin{bmatrix}
\frac{1}{2}(u+v) \\
\frac{1}{2}(u-v) \\
\end{bmatrix}
\,\Rightarrow\quad J_1=\frac{dR}{dq}=
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -\frac{1}{2} \\
\end{bmatrix}\\\\
ds^2=&a\,q'^T\,J_1^T\,\eta\,J_1\,q'=a\,du\,dv
\end{align*}
where
$\eta= \begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}\\\\$
Step II)
\begin{align*}
a&\mapsto {\it r_s}\,{{\rm e}^{-{\frac {r}{{\it r_s}}}}}{{\rm e}^{1/2\,{\frac {u-v
}{{\it r_s}}}}}{r}^{-1}
\\\\
ds^2&=a\,du\,dv={{\it du}}^{2}{\it r_s}\,{{\rm e}^{-1/2\,{\frac {2\,r-u+v}{{\it r_s}}}}}
{r}^{-1}-{{\it dv}}^{2}{{\rm e}^{1/2\,{\frac {2\,r-u+v}{{\it r_s}}}}}r{
{\it r_s}}^{-1}
\end{align*}
Step III)
\begin{align*}
R & =
\begin{bmatrix}
-\exp\left(\frac{u}{2\,r_s}\right) \\
\exp\left(-\frac{v}{2\,r_s}\right) \\
\end{bmatrix}\,,\Rightarrow\quad
J_2=\frac{dR}{dq}=\begin{bmatrix}
-\frac{2\,r_s}{\exp\left(\frac{u}{2\,r_s}\right)} & 0 \\
& -\frac{2\,r_s}{\exp\left(-\frac{v}{2\,r_s}\right)} \\
\end{bmatrix}\\\\
ds^2=&q'^T\,J_2^T\,J_1^T\,g\,J_1\,J_2\,q'=
\frac{4\,r_s^3\,\exp\left(-\frac{r}{r_s}\right)}{r}\,du\,dv
\end{align*}
Step IV
\begin{align*}
R & =
\begin{bmatrix}
u-v \\
u+v \\
\end{bmatrix}\,,\Rightarrow\quad
J_3=\frac{dR}{dq}=\begin{bmatrix}
1 & -1 \\
1 & 1 \\
\end{bmatrix}\\\\
ds^2=&q'^T\,J_3^T\,J_2^T\,J_1^T\,g\,J_1\,J_2\,J_3\,q' =
\frac{4\,r_s^3\,\exp\left(-\frac{r}{r_s}\right)}{r}\left(
du^2-dv^2 \right)
\end{align*}