From schaum's vector analysis:
I project the differential area $dS$ of the plane onto the $xy$ plane, then
$dxdy = dS (|\hat n. \hat k|)$ Where $\hat n$ is the normal vector to $dS$
then $dS = \frac{dxdy}{ |\hat n. \hat k| }$
$ \hat n = \frac{ \nabla S}{ | \nabla S | } = \frac{2}{3} \hat i + \frac {1}{3} \hat j + \frac{2}{3} \hat k$
$\nabla\times\vec F = 3 \hat i - \hat j - 2 \hat k$
$(\nabla\times\vec F) . \hat n = \frac{1}{3} $
$dS = \frac{3}{2} dxdy$
Then, $\iint_S \nabla\times\vec F \cdot\ \hat n $ $dS$ = $ \frac{1}{2} \iint_S dxdy = \frac{1}{2} \iint_0^2 dxdy$ = $ \frac{1}{2} \int_0^1 2 dx = 1$
Now this is is when the boundaries are $x=0, x=1, y=0, y=2$
But when the boundaries are $x=0 , y=0, z =0$ , as follows:
It's all the same steps except for:
$\iint_S \nabla\times\vec F \cdot\ \hat n = \frac{1}{2} \iint_S dxdy = \frac{1}{2} \iint_0^{6-2x} dxdy = \frac {1}{2} \int_0^3 6-2x dx = \frac {9}{2}$
Now what I don't understand is: In the first part, we treated $y$ as changing independently of $x$ from $y=0$ to $y=2$
In the second part, we treated $y$ as dependent on $x$ by the function $y=6-2x$ and integrated from $y=0$ to $y=6-2x$
Why is that? When do we substitute $y$ in as a function of $x$ and integrate like in the second part, and when not to? The problem is I cannot visualize 3 planes intersecting each other and visualize $S$ , so how can I understand it?
In the first problem, you are integrating the surface of the plane $\mathcal{P}: 2x+y+2z=6$ over the rectangle $0\leq x\leq 1, 0\leq y\leq 2$. To parametrize the rectangle, simply use $x,y$ as parameters and integrate as you did.
For the second problem, you are integrating the surface of $\mathcal{P}$ above the $xy-$plane. Imagine the plane $\mathcal{P}$ as a sheet of paper slicing into the $xy-$plane. The sheet creates a shadow over the $xy-$plane, which will be in the shape of a triangle. To see this, let $z=0$ to see what the plane looks like in the $xy-$plane. We'll get $$2x+y=6$$ Now, this is the line $y=6-2x$ which, when bounded by the lines $x=0$ and $y=0$, gives us a triangle. The triangle is a simple region and can be parametrized, for example, by $$0\leq x\leq 3$$ $$0\leq y\leq 6-2x$$ which explains the solution to the second problem.
To address your point about $y$ being "independent" from $x$ in the first integral and not the second, it's because the first region is a rectangle, which can be parametrized by all constant bounds (this is the easiest case). However, for more general regions (like the triangle in problem $2$), you cannot express them as $$a\leq x\leq b$$ $$c\leq y \leq d$$ but for example $$a\leq x\leq b$$ $$f(x)\leq y\leq g(x)$$ where $y$ is bounded by two continuous functions of $x$. In this problem, those two functions are $f(x)=0$ and $g(x)=6-2x$.