@MoisheKohan Do you know if there’s a counterexample with finitely generated groups? Surfaces of finite type feel a bit more tame. Although if you use free groups you might be able to get a planar/handle mismatch...

I should note that this general construction comes up often in very natural situations. At least, it does with groups I care about in geometry and topology.

I’d be curious to hear the algebro-geometric motivation behind this question

@HJRW Are you referring to the HNN construction? In that example, if you take the relation $a^t=ab$ and the group generated by all letters involved, you don’t get a proper subgroup. This never happens in OP’s situation.

It feels as though no such group exists. Since every finitely-generated subgroup is both proper and free by assumption, how could the group have any nontrivial relations?

Just as an aside, the smallest finite quotient for the full surface braid group on any number of strands (for $g>0$) is $\mathbb{Z}/(2)$; the “classical” generators collapse to an order two element in the abelianization.

I believe that Lawrence introduced this representation in 1990; Krammer cites Commun. Math. Phys. 135 (1990), 141-191.

You may be interested in the Alexander method, detailed in chapter 2 of the Primer on Mapping Class Groups. I believe Proposition 2.6 is the right reference.

I’m confused about your last sentence. There is no canonical Alexander system on a surface, but surely that doesn’t invalidate its usefulness as a tool. And, in fact, there is a nice useful bijection between mapping classes and Alexander systems (fixing a particular reference system).

You might be interested in this MO question and this partial answer (section 6) as referenced in the comments there.

As mentioned in your later question, the étale fundamental group acts on $Y$ by automorphisms of the étale morphism. In order for this action on $Y$ to give an automorphism of the cover, should fibers be preserved?

You should be heavily analogizing with algebraic topology at every step. The definitions given by Denis Nardin are equivalent because the fiber, seen as a $\pi_1$-set, has the same automorphism group as the cover does — namely, the deck group.