You're on the right track, but be careful about the Jacobian. It looks like if you proceed with the $u,v$ you chose and $J=-1/3$, then may wind up with a negative answer if not careful. Correct magnitude, wrong sign. Why? This is related to properties of the Jacobian. When negative, it will give you a negatively oriented region. This article spells it out in detail. In practice, just take the absolute value of J when necessary.. Everything else you wrote looks good. Let's compute the integral out using your $u$ and $v$ but using the new bounds, defined in terms of $u$ and $v$.
$$u=y-2x\ \ \ \ \ and \ \ \ \ \ v=x+y$$
implies
$$x=\frac{v-u}{3}\ \ \ \ \ and \ \ \ \ \ y=\frac{u+2v}{3}.$$
This gives
$$J = = \frac{1}{9}det\begin{bmatrix} -1 & 1 \\ 1 & 2 \end{bmatrix}=-\frac{1}{3}.$$
$$|J| = \frac{1}{3}$$
The original bounds for $y$ are $y=2x-2$ to $y=2x+2$. Substituting the above equations gives new bounds $v=u-3$ and $v=u+3$. Similarly for $x=-1$ to $x=1$, we get $u=-2$ to $u=2$.
Putting everything together, and moving the constant Jacobian to the front, we get
$$\frac{1}{3}\int_{-2}^2\int_{u-3}^{u+3}\ u^2 v^2 dv\ du$$
$$\frac{1}{9}\int_{-2}^2\ u^2 \{(u+3)^3 - (u-3)^3\}\ du$$
This doesn't look pretty, but think about Pascal's triangle for a bit, and you'll see that half of the terms inside the braces cancel, while the other half sum together, resulting in
$$\frac{2}{9}\int_{-2}^2\ u^2 (9u^2+27)\ du$$
$$2\int_{-2}^2\ u^4+3u^2\ du$$
This is an even integral, so
$$4\int_{0}^2\ u^4+3u^2\ du$$
$$4\left( \frac{u^5}{5}+u^3 \right)\bigg|_0^2$$
$$-4\left( \frac{32}{5}+8 \right) = 57.6$$
This matches Wolfram's output.
