If you are trying to prove the independence of $\overline X$ and $S^2$ via change of variables, suggest you using an orthogonal transformation for ease of calculation. The result can be proved without finding the joint pdf of $(\overline X,S^2)$ and showing they are independent because the joint pdf factors as the product of two marginals.
Consider the transformation $$(X_1,X_2,\ldots,X_n)\to(Y_1,Y_2,\ldots,Y_n)$$ such that $$\begin{pmatrix}Y_1\Y_2\\vdots\Y_n\end{pmatrix}=Q\begin{pmatrix}X_1\X_2\\vdots\X_n\end{pmatrix}$$
, where $Q$ is an $n\times n$ orthogonal matrix with the first row $$\left(\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}},\ldots,\frac{1}{\sqrt{n}}\right)$$
Then, $$Y_1=\frac{1}{\sqrt{n}}\sum_{i=1}^n X_i=\sqrt{n}\overline X \quad\text{ and }\quad\sum_{i=1}^n Y_i^2=\sum_{i=1}^n X_i^2$$
Clearly, $$(X_1,X_2,\ldots,X_n)\in\mathbb R^n\implies (Y_1,Y_2,\ldots,Y_n)\in\mathbb R^n $$
The absolute value of the Jacobian determinant is $$|J|=\frac{1}{|\det Q|}=1$$
Further,
\begin{align}
\sum_{i=1}^n (x_i-\mu)^2&=\sum_{i=1}^n x_i^2-2n\bar x\mu+n\mu^2
\&=\sum_{i=1}^n y_i^2-2\sqrt{n}y_1\mu+n\mu^2
\&=(y_1-\sqrt{n}\mu)^2+\sum_{i=2}^n y_i^2
\end{align}
So joint pdf of $(Y_1,Y_2,\ldots,Y_n)$ is of the form
\begin{align}
f_{Y_1,\ldots,Y_n}(y_1,\ldots,y_n)&=\frac{1}{(\sigma\sqrt{2\pi})^n}\exp\left[-\frac{1}{2\sigma^2}\left{(y_1-\sqrt{n}\mu)^2+\sum_{i=2}^n y_i^2\right}\right]
\&=\frac{1}{\sigma\sqrt{2\pi}}\exp\left[-\frac{1}{2\sigma^2}(y_1-\sqrt{n}\mu)^2\right],\prod_{j=2}^n \left{ \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{y_j^2}{2\sigma^2}\right)\right}
\end{align}
It is now clear that $Y_1,Y_2,\ldots,Y_n$ are independently distributed with
$$Y_1\sim\mathcal N(\sqrt{n}\mu,\sigma^2)\quad \text{ and }\quad Y_j\sim\mathcal N(0,\sigma^2),,\quad j=2,3,\ldots,n$$
While we get the distribution of $\overline X$ from $Y_1$, we get the distribution of $S^2$ from $Y_2,Y_3,\ldots,Y_n$.
$\overline X$ and $S^2$ are independently distributed precisely because $Y_1$ is independent of $Y_2,\ldots,Y_n$.
Noting that
\begin{align}
\sum_{i=2}^n Y_i^2&=\sum_{i=1}^n Y_i^2-Y_1^2
\&=\sum_{i=1}^n X_i^2-n\overline X^2
\&=\sum_{i=1}^n (X_i-\overline X)^2
\&=(n-1)S^2
\end{align}
, we have $$\frac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}$$
And from $Y_1$ we already had $$\overline X\sim \mathcal N\left(\mu,\frac{\sigma^2}{n}\right)$$
