Let $X,Y$ be independent normal r.v's each with parameters $\mu=0$, $\sigma^2=1$. Let us calculate the joint distribution of $(U,V)=(X+Y,X-Y)$. Here $$g(x,y)=(x+y,x-y)=(u,v),$$ and
$$g^{-1}(u,v)=(\frac{u+v}{2},\frac{u-v}{2}).$$ The Jacobian in this simple case does not depend on $(u,v)$ (that is, it is constant), and is
$$J_{g^{-1}}(u,v)=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \\ \end{pmatrix}$$
and....
The rest of the solution does not matter as I understand it. The problem involves using theorem 12.7 in Probability Essentials by Jean Jacod to find the join distribution for $X,Y$. I do not understand why the function is given by $g$ and how the inverse to $g$is given in such a way to thus find the Jacobian. Would someone please explain.
You have been told that random vector $\def\<{\langle}\def\>{\rangle}\<U,V\>$ is related to random vector $\<X,Y\>$ by some vector function such that $\<U,V\>=\<X+Y,X-Y\>$. We feel a need to call this function something, so we may as well name it: $g$. So if $\<x,y\>$ are the values of $\<X,Y\>$, then the values of $\<U,V\>$ will be $\<u,v\>=g(x,y)$ where: $$g(x,y)=\<x+y,x-y\>$$
Now, how are $\<X,Y\>$ related to $\<U,V\>$? Well, if $\<U,V\>=\<X+Y,X-Y\>$ then $\<X,Y\>=\<\tfrac 12(U+V),\tfrac 12(U-V)\>$ . This mapping is the inversion of $g$, so we have $\<x,y\>=g(u,v)$ where: $$g^{-1}(u,v)=\tfrac 12\<u+v,u-v\>$$
Now the Jacobian matrix is the matrix of partial differentials of the vector function's members with respect to the arguments, like so:$$\begin{align}J_{g^{-1}(u,v)}&=\begin{bmatrix}\dfrac{\partial g^{-1}_1(u,v)}{\partial u}& \dfrac{\partial g^{-1}_1(u,v)}{\partial v}\\\dfrac{\partial g^{-1}_2(u,v)}{\partial u}& \dfrac{\partial g^{-1}_2(u,v)}{\partial v}\end{bmatrix}\\[2ex]&=\begin{bmatrix}\dfrac{\partial \tfrac12(u+v)}{\partial u}& \dfrac{\partial \tfrac12(u+v)}{\partial v}\\\dfrac{\partial \tfrac12(u-v)}{\partial u}& \dfrac{\partial \tfrac12(u-v)}{\partial v}\end{bmatrix}\\[2ex]&=\begin{bmatrix}\tfrac12(1+0)& \tfrac12(0+1)\\\tfrac12(1-0)& \tfrac12(0-1)\end{bmatrix}\end{align}$$