applied probability

Let $f_X$ be density function of$X$, and let $f_Y$ be the density function of $Y$. Let $Z=X+Y$. Then $$f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\,dx.$$ Now let us apply this general formula to our particular case. We will have

                                                $f_Z(z)=0$  for $z\lt 0$,  and also for $z\ge 2$.

Now we deal with the interval from $0$ to $2$. Break this down into two cases (i) $0\lt z\le 1$ and (ii) $1\lt z\lt 2$.

(i) The product $f_X(x)f_Y(z-x)$ is $1$ in some places, and $0$ elsewhere. We want to make sure we avoid calling it $1$ when it is $0$. In order to have $f_Y(z-x)=1$, we need $z-x\ge 0$, that is, $x\le z$. So for (i), we will be integrating from $x=0$ to $x=z$. And easily $$\int_0^z 1\,dx=z.$$
Thus $f_Z(z)=z$ for $0\lt z\le 1$.

(ii) Suppose that $1\lt z\lt 2$. In order to have $f_Y(z-x)$ to be $1$, we need $z-x\le 1$, that is, we need $x\ge z-1$. So for (ii) we integrate from $z-1$ to $1$. Hence

$$\int_{z-1}^1 1\,dx=2-z.$$ Thus $f_Z(z)=2-z$ for $1\lt z\lt 2$.