I have this problem, I say that $x=z-y$, hence $F_z(z)=\iint\limits_D f(x,y)\ dx\ dy$, now I think that the limits for the $x$ integral would be from $0$ to $z-y$ and for $y$ would be from $0$ to $1$.
I am not sure if I can say that I need to analyze this when $0 <z <2$ and $-2 < z < 0$.
This is simply geometry... Rotate the $(x,y)$ domain you called the "shaded area", by $45°$ clockwise, now the abscissa axis corresponds to $z=x+y$. For every $|z|\leqslant1$, the height of the segment where the vertical line of abscissa $z$ intersects the domain is proportional to $f_Z(z)$. This segment is proportional to $[-|z|,+|z|]$, whose length is $2|z|$ for every $|z|\leqslant1$, hence $f_Z(z)=|z|\,\mathbf 1_{|z|\leqslant1}$. The CDF $F_Z$ follows by integration.
