I'm solving the below exercise.
Find the flux of the curl across $S$ in the direction $n$ of the field $F$, when $S : z = x^2 + 4y^2$ lying beneath the plane $z=1$, with the normal having a positive $k$-component, and $F = \langle y, -xz, xz^2\rangle$.
I applied two methods, but the results were not same.
When I let $S : g(x, y, z) = z - x^2 - 4y^2$, ($0 \leq z \leq 1$), then $n\,d\sigma = \nabla g / |\nabla g \cdot k| \,dx\,dy = \nabla g \,dx\,dy = \langle -2x, -8y, 1 \rangle\,dx\,dy$
If I define $x = u \cos v$, $y = (u/2) \sin v$, and $z = u^2$, ($0 \leq u \leq 1$, $0 \leq v \leq 2\pi$), then $n\,d\sigma = \langle -2u \cos v, -8u \sin v, 1\rangle u\,du\,dv$
However, when I let $ S : r(u, v) = (u \cos v, (u/2) \sin v, u^2)$, ($0 \leq u \leq 1$, $0 \leq v \leq 2\pi$), then $n\,d\sigma = (\partial r/ \partial u)\times(\partial r/\partial v)\,du\,dv= \langle -u^2 \cos v, -4u^2 \sin v, u/2\rangle\,du\,dv$
Here is the problem. The former is twice as large as the latter. What's wrong with the above?
The second one is correct, because $ d A $ is not $ u \, d u \, d v $. But also, you might want to try a different approach.
If $ x = u \cos v $ and $ y = u / 2 \sin v $, then $$ d A = d x \, d y = \left \lvert \matrix { \cos v & - u \sin v \\ 1 / 2 \sin v & u / 2 \cos v } \right \rvert \, d u \, d v = u / 2 \, d u \, d v \text . $$ This is your missing factor of $ 2 $.
You should also be careful with this method, since $ n \, d \sigma $ is not always $ \nabla g \, d A $ when the surface is given by $ g = 0 $; the more general formula is $ \nabla g / g _ z \, d A $; that is, you must divide by the $ z $ component of the gradient. But $ g _ z = 1 $ in this case, so that doesn't change anything; your method is correct as long as $ g $ takes the form $ z - f ( x , y ) $.
Now, you tagged this "stokes-theorem", and maybe this is part of a unit on Stokes's Theorem. In that case, maybe you're supposed to do this differently! Instead of integrating $ \nabla \times F \cdot n \, d \sigma $ across the surface, where $ \nabla \times F $ is the curl of $ F $, you could instead integrate $ F \cdot d r $ along the boundary of the surface. This is often quicker. (Or maybe you should do it both ways, and check that you get the same result.)