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-1$ there are $x_j\in X$ such that $$\big| f^j(z)-x_j\big| +\big|f(x_j)-f^{j+1}(z)\big|<\frac\epsilon m, \qquad (0\le j\le m-1).$$  Therefore  

$$d(z,z')\le \sum_{j=0}^{m-1}\Big\{\big|f^j(z)-x_j\big| +\big|f(x_j)-f^{j+1}(z)\big|\Big\}+\big|f^m(z)-z'\big|<2\epsilon,$$ 
which proves tha $d$ vanishes identically.

*All orbits of $f$ are dense* : Indeed, $f$ is conjugated with the irrational translation (modulo $1$) on $T_c:[0,1)\to[0,1)$ given by $T_c:x\mapsto \{x-c\}$, where $\{\cdot\}$ denotes the fractional part and $c:=\frac {\log 3}{\log 2}-1$. Consider the homeomorphism $h:[0,1)\to[1/2,1)$ defined by $h(x)=2^{x-1}$. Then it is easy to check that $h\circ f=T_c\circ h$.