Find all $u_0$ such that the recursive sequence defined by $u_{n+1}= 2\left| u_n \right| -1$ converges.
Let $f(x) = 2\left|x\right|-1$. If $(u_n)$ converges, it converges to a fixed point of $f$, that is to say $-\frac 13$ or $1$.
Furthermore, if $(u_n)$ converges to $l\in\{-\frac 13, 1\}$ , the mean value theorem yields, for some large enough $n_0$, $$\forall n\geq n_0, \displaystyle \left|\frac{u_{n+1}-l}{u_{n}-l} \right|=2$$
Since $\lim_{n\to \infty}(u_{n}-l)= 0$ it must be that $u_{n_1}=l\;\;$ for some $n_1\geq n_0$ which implies $u_n=l$ for all $n\geq n_1$.
Note that $u_0<-1$ or $u_0>1$ both lead to divergence of $(u_n)$
Therefore, finding the $u_0$ for which the sequence converges amounts to studying the following set $$X:= \{x\in[-1,1], \exists p\in \mathbb N, f^{(p)}(x)\in\{-\frac 13, 1\}\}$$
Is it possible to determine what this set contains exactly ? Certainly $X\cap (\mathbb R \setminus \mathbb Q)=\emptyset$.
Note that if $x\in I=[-1,1]$, then $f(x)\in I$, and as you have remarked, for $x$ to be in $X$, we need that $x\in I\cap \mathbb{Q}$.
Let $u_0=p/q$ in $X$, with $p$ prime to $q\geq 1$. Suppose that there exists a prime $l$, $l\not = 2,3$, such that $l$ divide $q$. Then if we set $u_1=\frac{p_1}{q_1}=\frac{2|p|-q}{q}$ ($p_1$ prime to $q_1$), we see that we have also that $l$ divide $q_1$. Hence as we want by a finite number of iteration to arrive at $1$ or $-1/3$, the only prime divisors of $q$ are possibly $2$ and $3$. The same argument works if $9$ divide $q$: we see that $9$ divide $q_1$. It remain to show that if $q=2^l$ or $q=3 2^l$, then $p/q$ is in $X$, If I am not wrong, it is easy.