There are several ways to find the CDF of $Z=X+Y$ with $X$ and $Y$ independent RVs. I want to use the general answer included in this question CDF of two variable to find the CDF when $X$ and $Y$ are continuous and uniform distributed over $[0,1]$.
In this case, I use the two densities to the be $1$. The result is $z^2/2$ when $0\le z\le1$ and this I know is correct. The result is $z-\frac{1}{2}$ when $1\le z \le2$ that is not correct, it should be: $1- \frac{1}{2}(2-z)^2$. Where is the mistake, in the cited solution or when I use it? Can you elaborate?
The answer included in CDF of two variable has been now edited and corrected by the author. The following is valid for a sum of two uniformly distributed continuous random variables:
$$\mathsf P(X+Y\leq z) = \begin{cases} 0 & : z< 0 \\[2ex] \int\limits_{0}^{z}\;\int\limits_{0}^{z-x} f_{X,Y}(x,y)\operatorname d y\operatorname d x & : 0 \leq z \leq 1 \\[2ex] \int\limits_{0}^{z-1}\;\int\limits_{0}^{1} f_{X,Y}(x,y)\operatorname d y\operatorname d x + \int\limits_{z-1}^{1}\;\int\limits_0^{z-x} f_{X,Y}(x,y)\operatorname d y\operatorname d x & : 1 < z \leq 2 \\[1ex] 1 & : 2 < z\end{cases}$$
where the outer integral is over $x$ variable. Let's explain the integration domain. It can be represented by $E\times F$ where both $E$ and $F$ are subsets of $[0,1]$. $F=[0,\min(z-x,1)]$ because $y\leq z-x\,\, \forall z$ and x.
To find $E$, it is suitable to split the unit interval depending on the value of z.
$0 \leq z \leq 1$:
$E=[0,z]\cup [z,1]$. However $x\in[z,1]$ does not fulfill the main $y\leq z-x$. Therefore $E=[0,z]$.
$1 < z \leq 2$:
$E=[0,z-1]\cup [z-1,1]$.
$x \in[0,z-1]\implies \min(z-x,1)=1 \implies F=[0,1]$
$x \in [z-1,1] \implies \min(z-x,1)=z-x \implies F=[0,z-x]$
Example picture for the sum of two integrals in the region $1 < z \leq 2$ (for $z =1.4$):