I would parameterize from the start, so on $C$ $-2\le z\le 2 $ and $x^2+y^2=4$. This is a cylinder of radius $2$ centred at the origin with height $4$.
So let $x=2\cos\phi,y=2\sin\phi $ where $ 0\le\phi\le 2\pi $ now we parametrise our surface $C$ as $\sigma (\phi,z)=(2\cos\phi,2\sin\phi,z)$, and now $F=(2z^2\sin\phi,0,0)$.
Here we let $ z$ increase from -2 to 2 so with this orientation, the normal $n=\sigma_\phi\times\sigma_z = \left| \begin{array}{ccc}
i & j & k \\
-2\sin\phi & 2\cos\phi & 0 \\
0 & 0 & 1 \end{array} \right|=(2\cos\phi,2\sin\phi,0)$
So we have $\int\int_C(\nabla\times F)\cdot dS=\int_0^{2\pi}\int_{-2}^2(\nabla\times F)\cdot(2\cos\phi,2\sin\phi,0)dzd\phi$
Now $ \nabla\times F=\left| \begin{array}{ccc}
i & j & k \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
yz^2 & 0 & 0 \end{array} \right|=(0,\frac{\partial}{\partial z}(yz^2),-z^2)=(0,2yz,-z^2)=(0,4z\sin\phi,-z^2)$
So $\iint_C(\nabla\times F) \cdot d\mathbf{S}=\int_0^{2\pi}\int_{-2}^2(0,4z\sin\phi,-z^2)\cdot(2\cos\phi,2\sin\phi,0)dz~d\phi
\\ =\int_0^{2\pi}\int_{-2}^2 8z\sin^2 \phi \, dz \, d\phi
=4 \int_0^{2\pi} (1-\cos2\phi) \, d\theta \; \int_{-2}^2 z \, dz =4 \color{#009900 }{[z^2]^2_{-2}} \; [\phi-\frac{1}{2}\sin 2\phi]^{2\pi}_0=\color{#009900 }{0}$
I took a different route because the "$\frac{\partial z}{\partial y}\times\frac{\partial z}{\partial x}$" didn't make sense to me. They are not vectors.
To do the $P$ integral I would also parametrize, but this time you are integrating over a cone.
Parametrising $P$, here we let $\sigma(\phi,z)=((4-z)\cos\phi,(4-z)\sin\phi,z)$, here
$n=\sigma_\phi\times\sigma_z =\left| \begin{array}{ccc}
i & j & k \\
(z-4)\sin\phi & (4-z)\cos\phi & 0 \\
-\cos\phi & -\sin\phi & 1 \end{array} \right|=((4-z)\cos\phi,(4-z)\sin\phi,4-z)$
so $\int\int_P (\nabla\times F)\cdot dS=\int_0^{2\pi}\int_2^4(0,2z(4-z)\sin\phi,-z^2)\cdot((4-z)\cos\phi,(4-z)\sin\phi,4-z)dzd\phi$
$=\int_0^{2\pi}\int_2^42z(4-z)^2\sin^2\phi-z^2(4-z)dz d\phi$
$=\int_0^{2\pi}\int_2^4z(4-z)^2(1-\cos 2\phi)-(4z^2-z^3)dzd\phi$
$=\frac{20}{3}(\phi-\frac{1}{2}\cos 2\phi|_0^{2\pi})-2\pi(\frac{44}{3})$
$=2\pi(\frac{20}{3}-\frac{44}{3})=-2\pi(\frac{24}{3})=-8(2)\pi=-16\pi$
Halelujah! To answer your new questions:
Cross producting two vectors produces a vector normal to both of them. Since $z$ is oriented positively, $\sigma_\phi\times\sigma_z$ produces the outward normal which is what we want, taking the other order gives us an inward normal.
In my belief you got lucky with your computation, since the integral you performed was on the base of the cone which is not actually part of the surface.
Addendum
I believe your main issue was with the reasoning for taking the normal to be $\sigma_\phi\times\sigma_z$ over $\sigma_z\times\sigma_\phi$. This may seem strange Please refer here.
