Let $S\subset \mathbb R^2$. If $S$ has the area $dxdy$ in $(x,y)$, then it will have the area $$|\det(x(u,v),y(u,v))|dudv$$ in $(u,v)$.
We commonly write $$dxdy=|\det(x(u,v),y(u,v)|dudv.$$
I'm not really sure how to interpret it. Would it be the area of $S$ in $(u,v)$ ? But in this case, $$|S|=\iint_Sdxdy=\iint_S|\det(x(u,v),y(u,v))|dudv,$$ so $dS=dxdy=|\det(x(u,v),y(u,v)|dudv$ ? But what does it really mean ?
You have in the $(x,y)$-plane the standard area measure ${\rm d}(x,y)$ and similarly in the "auxiliar" $(u,v)$-plane the standard area measure ${\rm d}(u,v)$. When you are given a (maybe complicated) domain $S$ in the $(x,y)$-plane and want to compute its area then you often use an essentially 1:1 parametrization of $S$ from an auxiliar domain $\hat S$ in the $(u,v)$-plane: $$\psi:\quad \hat S\to S,\qquad (u,v)\mapsto\bigl(x(u,v),y(u,v)\bigr)\ .$$ Such a parametrization will in general not be area conserving. In fact an arbitrary "area element" centered at some point $(u,v)\in\hat S$ will be mapped to a smaller or larger area element centered at the point $\bigl(x(u,v),y(u,v)\bigr)\in S$. The local area scaling factor turns out to be $$|J_\psi(u,v)|=\bigl|{\rm det}(d\psi(u,v))\bigr|\ .$$ This is often written as $${\rm d}(x,y)=\bigl|{\rm det}(d\psi(u,v))\bigr|\>{\rm d}(u,v)$$ and appears in the integral as $${\rm area}(S)=\int_S{\rm d}(x,y)=\int_{\hat S}\bigl|{\rm det}(d\psi(u,v))\bigr|\>{\rm d}(u,v)\ .$$ Note that I have just listed the usual formulas, I have proven nothing.