To solve this problem, I need to draw a picture. I strongly recommend that you do so also.
Note that since $0\lt x\lt 1$, we have $0\lt y\lt 2$.
So draw the rectangle with corners $(0,0)$, $(1,0)$, $(1,2)$, and $(0,2)$.
Draw the lines $y=x$ and $y=x+1$.
Our random variable lives in the rectangle, and between these two lines.
Now finding the (marginal) density function of $X$ is easy. We have to "integrate out" $y$. So $y$ will travel from $x$ to $x+1$.
In principle, finding the density function of $Y$ is also easy, we have to integrate out $x$.
But if you look at the picture, you can see that we will have to break up the integral into two parts.
If $0\lt y\le 1$, we will be integrating from $x=0$ to $x=y$. From $1\lt y\le 2$, we will be integrating from $x=y-1$ to $x=1$.
Best Answer
$$F_Z(z) = P(X+Y \leq z) = \int_{-\infty}^{\infty} \left[\int_{-\infty}^{z-x} f_{XY}(x,y)dy\right]dx$$
$$f_Z(z)=\frac{d}{dz}F_Z(z)=\int_{-\infty}^{\infty} \left[\frac{d}{dz}\int_{-\infty}^{z-x} f_{XY}(x,y)dy\right]dx$$
$$f_Z(z)=\int_{-\infty}^{\infty}f_{XY}(x,z-x)dx = \int_{-\infty}^{\infty}f_{X}(x)f_{Y}(z-x)dx$$