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 any $z$ and $z'$ in $Z$, and for any $\epsilon>0$, there is $m\in\mathbb N$ such that $|f^m(z)-z'|< \epsilon$. 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
$$\big| f^j(z)-x_j\big|<\frac\epsilon m, \qquad (0\le j\le m)$$
$$\big|f(x_j)-f^{j+1}(z)\big|<\frac\epsilon m, \qquad (0\le j\le m-1).$$ Therefore for $0\le j\le m-1$
$$\big|f(x_j)-x_{j+1}\big|\le \big|f(x_j)-f^{j+1}(z)\big|+\big|f^{j+1}(z)-x_{j+1}\big|\le \frac{2\epsilon}m$$
and
$$d(z,z')\le\big|z-x_0\big|+ \bigg(\sum_{j=0}^{m-1} \big|f(x_j)-x_{j+1}\big| \bigg) +\big|x_m-z'\big|\le 4\epsilon,$$
which proves tha $d$ vanishes identically.
All orbits of $f$ are dense : Indeed, having all orbits dense is a property invariant by conjugation, and $f$ is conjugated with 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 $h\circ f=T_c\circ h$.