Forget CDFs, in this case they only make things more difficult. Instead, start from the two points below:
If $X$ is uniform on $(0,1)$, then $2X$ is uniform on $(0,2)$ hence $2X-1$ is uniform on $(-1,1)$ hence $|2X-1|$ is uniform on $(0,1)$.
If $Y$ is uniform on $(0,1)$, then $1-Y$ is uniform on $(0,1)$ hence $2(1-Y)$ is uniform on $(0,2)$.
Putting these together yields the fact that $Z=2(1-|2X-1|)$ is uniform on $(0,2)$ for every $X$ uniform on $(0,1)$.
A (not very interesting) by-product is the CDF you know: $F_Z(z)=0$ for $z\leqslant0$, $F_Z(z)=\frac12z$ for $0\lt z\leqslant2$, $F_Z(z)=1$ for $z\gt2$.