Convolution of two Uniform random variables
Instead of trying to find appropriate $z$ values at the very beginning, note that $f_X(x)$ is zero unless $0\le x\le2$. Therefore $$\int_{-\infty}^{\infty}f_X(x)f_Y(z-x)\mathrm{d} x =\int_0^2f_X(x)f_Y(z-x)\mathrm{d} x =\frac12\int_0^2 f_Y(z-x)\mathrm{d} x\ .$$ You can now substitute $t=z-x$ to get $$\int_{z-2}^z f_Y(t)\mathrm{d} t$$ - see if you can take it from here.