Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ such that $$\begin{equation} \begin{cases} f(f(x)-f(y))=|f(x)-f(y)|, \\ f(1-f(x+2))=1-f(x) \end{cases} \end{equation}$$ for all $x, \; y \in \mathbb{R}$.
My work. $1)$ $x=y \Rightarrow f(0)=0$.
$2)$ $x=0 \Rightarrow f(1-f(2))=1$.
$3)$ If $x=1-f(2), \; y=0$ then from $ f(f(x)-f(y))=|f(x)-f(y)|$ it follows that $f(1)=1$.
$4)$ $x=1 \Rightarrow f(1-f(y))=|1-f(y)|$. If $y=x+2$ then from $ f(1-f(y))=|1-f(y)|$ it follows that $ f(1-f(x+2))=|1-f(x+2)|$. But $ f(1-f(x+2))=1-f(x)$. Then $1-f(x)=|1-f(x+2)|$. Then $f(x) \le 1$. Then $f(x+2) \le 1$. Then $1-f(x)=1-f(x+2) \Rightarrow f(x)=f(x+2)$. Then the function $f: \mathbb{R} \to \mathbb{R}$ is periodic with period $2$.
You have the conditions to satisfy of
$$f(f(x)-f(y))=|f(x)-f(y)| \tag{1}\label{eq1}$$
$$f(1-f(x+2))=1-f(x) \tag{2}\label{eq2}$$
As C.Park's comment stated, you're very close with what you've done so far. You've already determined that $f(0) = 0$, $f(1) = 1$ and $f(x) = f(x+2)$, i.e., it's periodic with a period of $2$, so I won't repeat your work here.
Next, you can use the given requirement that $f(x)$ be continuous, plus that $f(0) = 0$ and $f(1) = 1$, to be able to state that for all $z \in [0,1]$, there exists at least one $x \in [0,1]$ such that $f(x) = z$. Use this $x$, and $y = 0$, in \eqref{eq1} to get
$$f(f(x) - f(y)) = f(z - 0) = |f(x) - f(y)| = |z - 0| \tag{3}\label{eq3}$$
i.e.,
$$f(x) = x, \; \forall \; x \in [0,1] \tag{4}\label{eq4}$$
Next, let $x = 0$ and $y \in [0,1]$ in \eqref{eq1} again to get
$$f(f(x) - f(y)) = f(0 - y) = |f(x) - f(y)| = |0 - y| \tag{5}\label{eq5}$$
i.e., $f(x) = -x, \; \forall \; x \in [-1,0]$. Combining this with \eqref{eq4} gives
$$f(x) = |x|, \; \forall \; x \in [-1,1] \tag{6}\label{eq6}$$
Finally, you can use the periodicity aspect to get the graph is a "saw-tooth" type function, alternating between $0$ and $1$ at integral points. In particular, the function is
$$f(x) = |r - 1|, \; \forall \; x \in \mathbb{R}, \; \text{ where } \; x = 2n + r, \; \text{ with } \; n \in \mathbb{Z} \; \text{ and } \; 0 \le r \lt 2 \tag{7}\label{eq7}$$
You can confirm this equation satisfies \eqref{eq1} and \eqref{eq2}.