Evaluate $\int_c F \cdot dr$, where $F =\langle −y^2, x, z^2\rangle$, and C is the intersection of $y + z = 2$ and $x^2 + y^2 = 1$ positively oriented.
$$\operatorname{curl} F = 1+2y$$
I evaluated dS in two ways though and got different answers.
Firstly, I set $g(x,y)=z=2-y$ and calculated $dS = \sqrt{(dg/dx)^2+(dg/dy)^2+1} = \sqrt{2}$
Then I decided to parametrise the cylinder, express it as a vector function, and take the magnitude of the cross product:
$$r(z, \theta) = [\cos(\theta)i, \sin(\theta)j, zk]\\ r_z = [0,0,1]\\ r_\theta = [-\sin(\theta), \cos(\theta), 0]$$
The magnitude of the cross product of these two = $\sqrt{-1}$
Both calculations seem correct, I'm not sure what I've done wrong.
Your most fundamental mistake seems to be neglecting the vector nature of the right hand side of $$\oint_C\vec F\cdot d\vec r=\int\int_S\vec\nabla\times\vec F\cdot d^2\vec A$$ Let's do the left hand side first. On the cylinder we may parameterize $x=\cos\theta$, $y=\sin\theta$, and the intersection with the plane means $z=2-y=2-\sin\theta$. Then we can say that along the curve $C$, $$\vec r=\langle x,y,z\rangle=\langle\cos\theta,\sin\theta,2-\sin\theta\rangle$$ $$d\vec r=\langle-\sin\theta,\cos\theta,-\cos\theta\rangle\,d\theta$$ $$\vec F=\langle-\sin^2\theta,\cos\theta,4-4\sin\theta+\sin^2\theta\rangle$$ So that $$\begin{align}\oint_C\vec F\cdot d\vec r&=\int_0^{2\pi}\langle-\sin^2\theta,\cos\theta,4-4\sin\theta+\sin^2\theta\rangle\cdot\langle-\sin\theta,\cos\theta,-\cos\theta\rangle\,d\theta\\ &=\int_0^{2\pi}\left[\sin^3\theta+\cos^2\theta-4\cos\theta+4\sin\theta\cos\theta-\sin^2\theta\cos\theta\right]d\theta\\ &=2\pi\left[0+\frac12-0+0-0\right]=\pi\end{align}$$ Where we have used the average values of each term to evaluate the integral. If we wish to parameterize the part of the plane $y+z=2$ inside the cylinder $x^2+y^2=1$ we can parameterize $x$ and $y$ in term of polar coordinates and then set $z$ by the equation for the plane $$\vec r=\langle r\cos\theta,r\sin\theta,2-r\sin\theta\rangle$$ $$d\vec r=\langle\cos\theta,\sin\theta,-\sin\theta\rangle\,dr+\langle-r\sin\theta,r\cos\theta,-r\cos\theta\rangle\,d\theta$$ $$\begin{align}d^2\vec A&=\pm\langle\cos\theta,\sin\theta,-\sin\theta\rangle\,dr\times\langle-r\sin\theta,r\cos\theta,-r\cos\theta\rangle\,d\theta\\ &=\pm\langle0,r,r\rangle\,dr\,d\theta=\langle0,r,r\rangle\,dr\,d\theta\end{align}$$ The $z$-component pointing up consistent with our counterclockwise traversal of the boundary earlier. Here you don't want just the magnitude of the vector areal element, you want the vector areal element itself. Also you need $\vec F$ in terms of the parameterization used, so $$\vec F=\langle-y^2,x,z^2\rangle$$ $$\nabla\times\vec F=\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial y}\rangle\times\langle-y^2,x,z^2\rangle=\langle0,0,1+2y\rangle=\langle0,0,1+2r\sin\theta\rangle$$ Then $$\begin{align}\int\int_S\vec\nabla\times\vec F\cdot d^2\vec A&=\int_o^{2\pi}\int_0^1\langle0,0,1+2r\sin\theta\rangle\cdot\langle0,r,r\rangle\,dr\,d\theta\\ &=(2\pi)(1)(\frac12)=\pi\end{align}$$ Where again we have used average values to make quick work of the integrals. So that was your problem with your first parameterization. The issue with the parameterization along the cylinder was that firstly the cylinder isn't going to from a closed surface so you would need an extra surface parallel to the $xy$-plane to close off one end or the other. And secondly you don't want the scalar areal element, you need the vector areal element just as was the case for the closure of the curve being the plane. And thirdly you calculated the magnitude of the scalar areal element wrongly in any case. Magnitude of areal element depends on the surface and on the parameterization and both were different in this case.