Yet, even though all of this is cute, I'd prefer a more "classical" reference to the basic result for the case "$\mathbb{Q}$ vs $\mathbb{R}$".

(...) $n:=aq-cp$ yields $M(m/n)=p/q$, to the effect that, by specializing the Choi-Vaaler inequality to the case mentioned in Vesselin's last comment, we have: $$\left|M(x)-\frac{p}{q}\right|=|M(x)-M(m/n)|\ge \eta_\infty(A)^{-1}\cdot\left|x-\frac{m}{n}\right|\ge\frac{\eta_\infty(A)^{-1}}{|n|^{\mu-2\varepsilon}}\ge\frac{\kappa}{q^{\mu-2\varepsilon}}\ge\frac{1}{q^{\mu - \varepsilon}},$$ where $\kappa := \eta_\infty(A)^{-1} \cdot ((|a|+|c|) \cdot (M(x)+1))^{2\varepsilon - \mu}$ and $A$ is the matrix of $M$. The conclusion follows from the arbitrariness of $\varepsilon$.

Thanks, Vesselin, for your explanations. Finally, I seem to get the point, but let me try to write down some further details: It suffices to prove that $\mu(M(x))\le\mu:=\mu(x)$. If $\mu=\infty$ or $x$ is rational, this is trivial, so assume $\mu < \infty$ and $x\notin\mathbb{Q}$, and pick $\varepsilon > 0$. We'd like to prove that $|M(x)-p/q|\ge q^{-\mu+\varepsilon}$ for all but finitely many reduced pairs $(p,q)\in\mathbb{Z}\times\mathbb{N}^+$ with $|qM(x)-p|<1$. For, pick $\varepsilon>0$ and a reduced pair $(p,q)$ with $q$ large enough. Then, putting $m := -bq+dp$ and (...)

I looked closer to Choi and Vaaler's results and Cassels' corollary on p. 11 of his book, and this morning, on a second thought, I don't see the relation with my questions. Those results do not apply to the Liouville-Roth irrationality exponent, but to the alternative measure of irrationality that we get by replacing $n^s$ with $s^{-1}n^2$ in the OP (which is relevant, in particular, in the study of badly approximable numbers). What do I miss?

Thank you, Vesselin, all the more that Choi and Vaaler's paper points to Cassels' An introduction to diophantine approximation (Ch. I, Sect. 3, p. 11) for the basic case of ${\rm PGL}(2,\mathbb{Q})$ and approximations to reals by rational numbers.

I made a mess with the table, but you are still right: The construction as it is doesn't work. Is there any hope to fix it?

Thanks for your clarifications, but you will agree that "common usage" has a very relative meaning: In France, e.g., people are known for speaking mathematics in a quite different way than elsewhere, regardless of their area of expertise; also, you mentioned Rolfsen, and Rolfsen, for an instance, does regard the fundamental grp of $\mathbb P^2$ as a surface group (see p. 4 in the survey linked in the OP). Lastly, it's perhaps worth remarking that, in their book, Rhemtulla (you seem to have mistyped his name) and Mura use the expression "orderable group" in a different sense than in the OP.