Line element in Kruskal coordinates I try to calculate the line element in Kruskal coordinates, these coordinates use the Schwarzschild coordinates but replace $t$ and $r$ by two new variables.
$$
T = \sqrt{\frac{r}{2GM} - 1} \ e^{r/4GM} \sinh \left( \frac{t}{4GM} \right) \\
X = \sqrt{\frac{r}{2GM} - 1} \ e^{r/4GM} \cosh \left( \frac{t}{4GM} \right)
$$
Wikipedia shows the result of the line element.
$$ 
ds^2 = \frac{32 G^3M^3}{r} e^{-r/2GM} (-dT^2 + dX^2) + r^2d\Omega^2
$$
I tried to calculate the metric tensor using $ds^2 = g_{ij} \ dx^i dx^j$. As $T$ and $X$ show no dependence in $\theta$ and $\phi$, the $d\Omega$ seems to make sense, but the calculation of the first component of $g$ was not working.
$$
g_{tt} = J^TJ = \frac{\partial T}{\partial t} \frac{\partial T}{\partial t} + \frac{\partial X}{\partial t} \frac{\partial X}{\partial t}\\
= \frac{1}{32} \left( \frac{r}{GM} - 2 \right) \frac{ e^{\frac{1}{2} \frac{r}{GM}}}{G^2M^2} \left( \cosh^2 \left( \frac{t}{4GM} \right) + \sinh^2 \left( \frac{t}{4GM} \right) \right)
$$ 
Is this the right way to compute the line elements? 
What would be better way to calculate the line elements (maybe starting with the Schwarzschild-coordinates)?

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*}
