First on $\mathbb R^2$, the space of the 2-form has dimension 1 and thus $$\mathrm{d}x\mathrm{d}y=\alpha \mathrm{d}u\mathrm{d}v.$$ Let find $a$. I denote $\partial _u$ for $\frac{\partial }{\partial u}$ and $x_u$ for $\frac{\partial x}{\partial u}$.
We have that $\mathrm{d}x=a\mathrm{d}u+b\mathrm{d}y$ and thus $x_u=\mathrm{d}x(\partial u)=a$ and $x_v=\mathrm{d}x(\partial _v)=b$. Therefore $$\mathrm{d}x=x_u\mathrm{d}u+x_v\mathrm{d}v$$ and $$\mathrm{d}y=y_u\mathrm{d}u+y_v\mathrm{d}v.$$
Now $$\mathrm{d}x\wedge \mathrm{d}y=(x_u\mathrm{d}u+x_v\mathrm{d}v)\wedge (y_v\mathrm{d}u+y_v\mathrm{d}v)=\\=(x_uy_v-x_vy_y)\mathrm{d}u\wedge \mathrm{d}v=\det J_{u,v}(x(u,v),y(u,v))\mathrm{d}u\wedge \mathrm{d}v.$$
Q1) Could someone explain way my proof is not rigorous ? Because I used exactly definition and theorem I have in my course of differentiable manifold, but my teacher said that what I did is not formal.
Q2) Supposed I proved that $$\mathrm{d}x\wedge \mathrm{d}y=\det J_{u,v}(x,y)\mathrm{d}u\wedge \mathrm{d}v$$
can I conclude directly that $$\iint_U \mathrm{d}x\mathrm{d}y=\iint_V |\det J_{u,v}(x,y)|\mathrm{d}u\mathrm{d}v \ \ ?$$ (if of course I have the right domain $U$ and $V$).
Q3) In fact, before I had to write $\mathrm{d}x\wedge \mathrm{d}y$ and $\mathrm{d}u\wedge \mathrm{d}v$ that disappeared in the integral. I don't really why we don't write it in the integral. May be $\mathrm{d}x\mathrm{d}y=|\det J_{u,v}(x,y)|\mathrm{d}u\mathrm{d}v$ rather than $\mathrm{d}x\wedge \mathrm{d}y=|\det J_{u,v}(x,y)|\mathrm{d}u\wedge \mathrm{d}v$ ? But if so, I know that $$\mathrm{d}x\mathrm{d}y=-\mathrm{d}y\mathrm{d}v$$ but $$\iint \mathrm{d}x\mathrm{d}y=\iint \mathrm{d}u\mathrm{d}v,$$ so since in the integral it commute, it should have an other reason.