Normally when we do a change of variables with multiple integrals (scalar valued function) we take the standard approach: $$\int F(x,y) \cdot dx \cdot dy =\int F(g(u,v)) \cdot Jac(g) \cdot du\cdot dv$$
If I instead do u substitution with $$dx = \partial_u x\cdot du + \partial_v x\cdot dv$$ and $$dy = \partial_u y\cdot du + \partial_v y\cdot dv$$
Then dx*dy comes out nasty. Is there any way I can simplify it to get the standard approach?
You need to think of $dx \, dy$ as really being the "wedge product" $dx \wedge dy$ (read about it here). Then if you do your substitution, the Jacobian formula should pop out. I'm going to use the subscript notation for derivatives to make things easier to read: $x_u = \frac{\partial x}{\partial u}$, etc...
So first just substitute as you suggest for $dx$ and $dy$, and then distribute:
$$ \begin {align*} dx \wedge dy &= \left( x_u \, du + x_v \, dv \right) \wedge \left( y_u \, du + y_v \, dv \right) \\ &= x_uy_u \, du \wedge du \, + \, x_vy_v dv \wedge dv \, + \, x_uy_v du \wedge dv \, + \, x_v y_u dv \wedge du \end {align*} $$
Since the wedge product is skew-symmetric, $du \wedge du =0$ and $dv \wedge dv = 0$, and also $dv \wedge du = - du \wedge dv$. So this all simplifies down to
$$ dx \wedge dy = \left( x_uy_v - x_vy_u \right) du \wedge dv $$
The coefficient $x_uy_v-x_vy_u$ is exactly the Jacobian determinant in the formula which you call the "standard approach".