This question comes from MIT OCW 6.041.
I don't understand how did I interpret the question wrong. The solution states that both $Y$ and $Z$ are stated (in the question) in terms of conditional, but I did not at all detect that when I was interpreting the question.
To me, $Y$ and $Z$ were just ordinary random variable. So a diagram of $y$ vs $x$ would be like the $f_X(x)$ vs $x$ diagram, except that the $y$ axis is $y$ between $0$ and $80$ (because it ranges from $0$ to $2x$).
As for $Z$, I thought $E[Z]$ is just a straight forward $\int^{40}_0\int^{2x}_x \frac{(y-x)}{1600} dy dx$ (because $E[g(X,Y)]=\int \int g(x,y)f_{X,Y} (x,y) dx dy$).
So why are $Y$ and $Z$ interpreted that way (in terms of conditional)?
The problem is saying that $Y\sim \mathrm{Unif}(0,2t)$, that is, the distribution on $Y$ depends on a parameter $t$, and $t$ belongs to the image of $X$, this is a clumsy way to say that
$$ \Pr [Y<c|X=t]=\int_{0}^c \mathbf{1}_{[0,2t]}(x)\frac1{2t}\mathop{}\!d x=\begin{cases} \frac{c}{2t},&c\in[0,2t]\\ 0,& c<0\\ 1,&c>2t \end{cases} $$ Or, equivalently, that $f_{Y|X}(c,t)=\frac1{2t}\mathbf{1}_{[0,2t]}(c)$.