One gets the trivial semi-distance. Let $d$ denote the semi-distance on $Z$ that defines $d_\sim$ on $Z/\sim$, that is $d_\sim([z],[z']):=d(z,z')$ for all $z$ and $z'$ in $Z$. Thus $d(z,z')\le |z-z'|$ for all $z$ and $z'$ in $Z$ and $d(x,f(x))=0$ for all $x\in X$.
It is convenient to extend the definition of $f$ to a self-map $[1/2,1)\to[1/2,1)$, still denoted $f$: $$f(x)=\begin{cases}\frac{4x}3 &\text{if}& 0\le x<\frac34\\ \frac{2x}3& \text{if}& \frac34\le x <1.\end{cases}$$
The key point is that all orbits of $f$ are dense. So for every $z$ and $z'$ in $Z$, and for every $\epsilon>0$, there is $m\in\mathbb N$ such that $\lvert f^m(z)-z'\rvert< \epsilon$. Since $X$ is also dense, and since $f$ is everywhere (right) continuous, for all indices $ 0\le j\le m$ there are $x_j\in X$ such that
\begin{align*}
\bigl\lvert f^j(z)-x_j\bigr\rvert<\frac\epsilon m, & \qquad (0\le j\le m) \\
\bigl\lvert f(x_j)-f^{j+1}(z)\bigr\rvert<\frac\epsilon m, & \qquad (0\le j\le m-1).
\end{align*}
Therefore for $0\le j\le m-1$
$$\bigl\lvert f(x_j)-x_{j+1}\bigr\rvert \le \bigl\lvert f(x_j)-f^{j+1}(z)\bigr\rvert+\bigl\lvert f^{j+1}(z)-x_{j+1}\bigr\rvert\le \frac{2\epsilon}m$$
and
$$d(z,z')\le\bigl\lvert z-x_0\bigr\rvert+ \bigg(\sum_{j=0}^{m-1} \bigl\lvert f(x_j)-x_{j+1}\bigr\rvert \bigg) +\big|x_m-z'\big|\le 4\epsilon,$$
which proves that $d$ vanishes identically.
All orbits of $f$ are dense: Indeed, having all orbits dense is a property invariant by conjugation, and $f$ is conjugate to the irrational translation (modulo $1$) on $[0,1)\sim\mathbb{R/Z }$ given by $T_c:x\mapsto {x -c}$, where $c:=\frac {\log 3}{\log 2}-1$ (all orbits of a translation $T_c$ are dense iff $c$ is irrational).
Consider the homeomorphism $h:[0,1)\to[1/2,1)$ defined by $h(x)=2^{x-1}$. Then it is easy to check that $h\circ f=T_c\circ h$$h\circ T_c=f \circ h$.
