I have a function defined as $g(x_1,x_2,x_3,u) = (x_1,x_2 - u, x_3,x_2 + u)$ that I am trying to find the Jacobian. In this case, the Jacobian is 1 since we obtain
$$\left| \begin{array}{cccc} dx_1/dx_1 & dx_1/dx_2 & dx_1/dx_3 & dx_1/du \\ d(x_2-u)/dx_1 & d(x_2 - u)/dx_2 & d(x_2-u)/dx_3 & d(x_2 -u)/du \\ dx_3/dx_1 & dx_3/dx_2 & dx_3/dx_3 & dx_3/du\\d(x_2+u)/dx_1 & d(x_2 + u)/dx_2 & d(x_2 + u)/dx_3 & d(x_2 +u)/du \end{array} \right| = \left| \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & -1 \\ 0 & 0& 1 & 0 \\ 0 & 1& 0&1\end{array} \right| = 1$$
However, if I define $h(x_1,x_2,x_3,u) = (x_1,x_3,x_2 - u, x_2 + u)$ then the Jacobian becomes
$$\left| \begin{array}{cccc} dx_1/dx_1 & dx_1/dx_2 & dx_1/dx_3 & dx_1/du \\ dx_3/dx_1 & dx_3/dx_2 & dx_3/dx_3 & dx_3/du\\d(x_2-u)/dx_1 & d(x_2 - u)/dx_2 & d(x_2-u)/dx_3 & d(x_2 -u)/du \\ d(x_2+u)/dx_1 & d(x_2 + u)/dx_2 & d(x_2 + u)/dx_3 & d(x_2 +u)/du \end{array} \right| = \left| \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0& 1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 1& 0&1\end{array} \right| = 0$$
Why does is the Jacobian so sensitive to this slight change in ordering? What does a Jacobian of 0 mean?
As Daniel pointed out, your evaluation of the determinants is in error.
$\begin{vmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & -1 \\ 0 & 0& 1 & 0 \\ 0 & 1& 0&1\end{vmatrix} = 1\begin{vmatrix} 1 & 0 & -1 \\ 0& 1 & 0 \\ 1& 0&1\end{vmatrix}-0\begin{vmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0&1\end{vmatrix}+0\begin{vmatrix} 0 & 1 & -1 \\ 0 & 0& 0 \\ 0 & 1& 1\end{vmatrix}-0\begin{vmatrix} 0 & 1 & 0 \\ 0 & 0& 1 \\ 0 & 1& 0\end{vmatrix} = 2$
And similarly the other equals $-2$.