Proof Strategy - Without Stokes's Theorem - Calculate $\iint_S \operatorname{curl} \mathbf{F} \cdot\; d\mathbf{S}$ for $\mathbf{F} = yz^2\mathbf{i}$ 
2013 10C. Question: Consider the bounded surface S that is the union of $x^2 + y^2 = 4$ for $−2 \le z \le 2$
  and $(4 − z)^2 = x^2 + y^2 $ for $2 \le z \le 4.$ Sketch the surface.
  Use suitable parametrisations for the two parts of S to verify Stokes’s Theorem for 
  $\mathbf{F} = (yz^2,0,0)$.

Herein, I'm asking only about proof strategy for Without Stokes's Theorem - Calculate $\iint_S \operatorname{curl} \mathbf{F} \cdot\; d\mathbf{S}$ for $\mathbf{F} = yz^2\mathbf{i}$ - 2013 10C, so omit irrelative calculations here. As I do there, denote the $2 \le z \le 4$ cone P, and the $-2 \le z \le 2$ cylinder C. 
User ellya's solution: $\bbox[3px,border:2px solid gray]{ \text{ 1st, do surface integral over cylinder  C} }$ Parameterize from the start. 
So let  $x=2\cos\phi \text{ and } y=2\sin\phi $,  where $ 0\le\phi\le 2\pi $ and $-2 \le z \le 2$. 
Now parameterise our surface $C$ as $\mathbf{r} (\phi,z)=(2\cos\phi,2\sin\phi,z) \implies \mathbf{ F \, (r} (\phi,z) \, ) =(2z^2\sin\phi,0,0)$.

$1.$ Why parameterise with $x = ...\cos\theta, y = ...\sin\theta$?
  Why not reverse them, say $y = ...\cos\theta, x = ...\sin\theta$? Would this still function?

With this orientation, the normal $\mathbf{n} =\mathbf{r}_{\phi}\times \mathbf{r}_z   =(2\cos\phi,2\sin\phi,0)$

$2.$ How would one determine that it's $\color{green}{(\partial_{\phi} \mathbf{r} \times \partial_{z} \mathbf{r}} )$ that 'produces the outward normal which is what we want', and not $\color{darkred}{(\partial_{z} \mathbf{r}  \times \partial_{\phi} \mathbf{r} )}$, for BOTH pieces C & P?
  I'm aware that everything must be positively oriented by convention.
  I'm not enquiring about a geometric or visual answer. 

$ \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) \quad \color{orangered}{ (*) }$
Substitute the parametrisation for C into $\color{orangered}{ (*) } : curl F   =(0,4z\sin\phi,-z^2)$.
So $\iint_C(\nabla\times \mathbf{F})\cdot d\mathbf{S}  = \int_0^{2\pi}\int_{-2}^2(0,4z\sin\phi,-z^2)\cdot(2\cos\phi,2\sin\phi,0) \,dz ~ d\phi    = ...$
$\bbox[3px,border:2px solid gray]{ \text{ 2nd and last, compute for cone $P$. } }$ 
The question hypothesises $4 - z = $ radius of the cone, so $z = 4 - r$. The cross-section of any cone is just a circle with radius $z$. So parameterise with 
$\mathbf{r} (\phi,z)=((4-z)\cos\phi,(4-z)\sin\phi,4-r)$ where $ 0\le\phi\le 2\pi $ and $2 \le z \le 4$.
Substitute the parametrisation for P into $\color{orangered}{ (*) } : curl F   =(0,2z(4-z)\sin\phi,-z^2)$.
$\mathbf{n} =\mathbf{r}_{\phi}\times \mathbf{r}_z = ((4-z)\cos\phi,(4-z)\sin\phi,4-z)$

$3,4$: Same questions $1, 2$, but for P now. 

 A: $1.$ Okay so first off we consider the cone, and the parametrisation $\sigma(\phi,z)=(\cos\phi,\sin\phi,z)$ is one way to parametrise, you could do $\sigma(\phi,z)=(\sin\phi,\cos\phi,z)$, this is fine, but it is just oriented in the opposite direction, i.e. in $2D$:
$(\cos\phi,\sin\phi)$ gives us a circle oriented anticlockwise.
but $(\sin\phi,\cos\phi)$ gives us a circle oriented clockwise.
So when it comes to finding normals we just choose the opposite one.
$2.$ Now on the subject of why we choose $\sigma_\phi\times\sigma_z$ over   $\sigma_z\times\sigma_\phi$ it really is best to think of this visually:
Imagine you have the $x$ and $y$ axis, if you take your index finger to be the $x$ axis and your middle finger to by the $y$ axis, and then you stick out your thumb so that all three are orthogonal to one another, then your thumb becomes the $z$ axis. (I think of all of this with my left hand) i.e $\underline{i}\times\underline{j}=\underline{k}$
If you then change the order, i.e. you swap $x$ and $y$ the cross product produces the $z$ axis in the negative direction. i.e. $\underline{j}\times\underline{i}=-\underline{k}$
Things get trickier when we consider $\phi$ and $z$, because $\color{red}{ \phi \text{ points in the direction tangent to the curved surface  } } $, and it points in the anticlockwise direction, (orthogonal to $z$) as if you are looking at the cylinder from a birds eye view (from above). Why? I explain in the end. 
See Determine Cross Product with Left Hand vs Right Hand. Let the $\phi$ direction be your index finger, and the $z$ axis be your thumb. the cross product produces a vector in the direction of your "middle finger", now imagine the back of your hand rests adjacent to the side of the cylinder, and you will realise that this new direction is the "outward" normal.
If we changed orientation, we would get the inward normal.
With regards to the cone, the answers $3,4$ are essentially identical.
Because if we were to switch the parametrization, we would be going around the cone in the opposite direction, hence reversing the orientation. 
And again the visual idea is the same, since we just imagine our "hand" is adjacent to the curved surface.
I hope that helps, but please feel free to ask more. 
why does $\phi$ point in the direction tangent to the curved surface?
When we consider a cylinder in $3D$ it is really just an infinite stack of circles one on top of the other, and we parameterize each circle with $(\cos\phi,\sin\phi)$ where $0\le\phi\le 2\pi$.
Now imagine we are directly above the cone looking down, what we will see is a circle (i.e. the image on the right), the circle has a fixed radius, and $\phi$ is the angle between the radius and a horizontal line that's coloured in yellow. As $\phi$ increases, we go around the circle anticlockwise.

Imagine you are standing on the edge of circle, always at a fixed distance from the centre. Then imagine you are running around it exactly at the rate that the angle increases.
You start where $\phi=0$ (on the diagram). Imagine you are about to run, the direction you are facing in is the direction of $\phi$, and at $\phi=0$ on the right hand diagram this direction is directly up. but more importantly this direction is tangent to the circle at that point, and as $\phi$ increases, you run, adjusting your direction accordingly so the the direction you face in is always tangent to the circle.
Now think about this on the cylinder, if you are running in a direction tangent to one of the circles, you are tangent to all of the circles (as they are stacked above each other). So you are in fact tangent to the curved surface.
A: Thanks to user ellya, I revamped ellya's sketch:

