PDF
We can first derive the PDF using convolution of two PDFs:
Case 1: If $0 \leq y \leq 1$, then $f_{X_1}(y - x_2) = 1$ if $ 0\leq x_2 \leq y$, and $f_{X_1}(y-x_2) = 0$ if $x_2 > y$. This means that
\begin{equation}
\int_0^1 f_{X_1}(y-x_2) dx_2 = \int_0^y 1 dx_2 = y
\end{equation}
Case 2: If $1 < y < 2$, then $f_{X_1}(y-x_2) = 1$ if $y-1\leq x_2 \leq 1$, and $f_{X_1}(y-x_2) = 0$ otherwise. So
\begin{equation}
\int_0^1 f_{X_1}(y-x_2) d x_2 = \int_{y-1}^1 1 dx_2 = 2-y
\end{equation}
Conclusion: $f_Y(y) = y$ if $0 \leq y \leq 1$ and $2-y$ if $1 \leq y \leq 2$. Otherwise, it is zero.
CDF
This means the CDF, which is defined as follows
\begin{equation}
F_Y(y) = \int_{-\infty}^y f_Y(y) dy
\end{equation}
Case 1: If $y < 0$: Clearly $F_Y(y) = 0$
Case 2: If $0 <y < 1$, then
\begin{equation}
F_Y(y) = \int_{0}^y f_Y(t) dt = \int_0^y t dt = \frac{y^2}{2}
\end{equation}
Case 3: If $1<y<2$, then
\begin{equation}
F_Y(y) = \int_{0}^y f_Y(t) dt = \int_0^1 t dt + \int_1^y 2-t dt =\frac{1}{2} -\dfrac{y^2-4y+3}{2}
\end{equation}
Case 4: If $y > 2$, then $F_Y(y) = 1$.