Per Bowditch, a group is relatively hyperbolic if it acts geometrically finitely on a proper geodesic Gromov hyperbolic space. A free product of two (or finitely many) finitely generated groups is well known to be relatively hyperbolic. It also acts on an associated BassSerre tree which is however locally infinite. My question: Is there a nice explicit description of a proper geodesic Gromov hyperbolic space on which the free product acts geometrically finitely?

$\begingroup$ So, how does the proof of relative hyperbolicity for free products work, if not by exhibiting an action? $\endgroup$ – ThiKu Mar 3 at 9:01

$\begingroup$ In his paper "Relatively Hyperbolic Groups" Bowditch proved that relative hyperbolicity (with the above definition) is equivalent to admitting an action on a (locally infinite) hyperbolic graph K such that the following condition hold. 1) All edge stabilizers are finite. 2) The number of orbits of edges is finite. 3) The graph K is fine, that is, for every n ∈ N, any edge of K is contained in finitely many circuits of length n. (Here circuit means a cycle without self– intersections). $\endgroup$ – Yellow Pig Mar 3 at 11:29

1$\begingroup$ But I wanted a very explicit description of the action on a proper hyperbolic metric space. It feels like there should be some action on a finite valence tree lurking... $\endgroup$ – Yellow Pig Mar 3 at 11:37

1$\begingroup$ @ThiKu probably because in the definition of relative hyperbolicity, no properness (of the space) is required. $\endgroup$ – YCor Mar 3 at 12:18

1$\begingroup$ @YellowPig, I think your intuition isn't right here. A proper action on a finitevalence tree would imply that your group is virtually free. $\endgroup$ – HJRW Mar 3 at 18:13
I think that the characterisation of relatively hyperbolic groups given in Groves and Manning's article Dehn filling in relatively hyperbolic groups answers your question. I give a few details:
Let $G:=\underset{1 \leq i \leq n}{\ast} A_i$ be a free product of $n$ finitely generated groups. For every $1 \leq i \leq n$, fix a finite generating set $S_i$ of $A_i$. Clearly, $S:= \bigcup\limits_{i=1}^n S_i$ is a finite generating set of $G$. The Cayley graph $\mathrm{Cayl}(G,S)$ is naturally a tree of spaces, the vertexspaces being Cayley graphs of $A_i$'s. Now, the idea is to glue "horoballs" on the vertexspaces. More precisely, consider $$X:= \left( \mathrm{Cayl}(G,S) \cup \bigcup_\limits{g \in G} \bigcup\limits_{i=1}^n \mathcal{H}(gA_i) \right) / \sim,$$ where the combinatorial horoball $\mathcal{H}(gA_i)$ over $\mathrm{Cayl}(A_i,S_i)$ is glued on $gA_i$.
The combinatorial horoball $\mathcal{H}(Y)$ over a graph $Y$ is defined as follows. The vertexset of $\mathcal{H}(Y)$ is $Y \times \mathbb{N}$. If $u$ and $v$ are two adjacent vertices of $Y$, connect $(u,0)$ and $(v,0)$ with an edge. Also, for every $k \geq 0$ and for every vertex $u \in Y$, connect $(u,k)$ and $(u,k+1)$ with an edge. Finally, for every $k \geq 0$, if $u,v \in Y$ are two vertices satisfying $d_Y(u,v) \leq 2^k$, connect $(u,k)$ and $(v,k)$ with an edge.
It turns out that $X$ is a proper hyperbolic space, and $G$ naturally acts on it.

$\begingroup$ This is a nice answer, but I think the OP was hoping for another hyperbolic space whose description is wellknown. Note that you can also glue ideal triangles rather than combinatorial horoballs, using Bowditch's construction. $\endgroup$ – M. Dus Mar 3 at 16:57

$\begingroup$ Gluing horoballs is a pretty well known operation, going back pretty far into the history of Gromov hyperbolic spaces. $\endgroup$ – Lee Mosher May 19 at 2:12
Too long for a comment. Here is another approach when the free factors are torsionfree abelian groups. Let $G=\mathbb{Z}^{d_1}*\mathbb{Z}^{d_2}...*\mathbb{Z}^{d_n}$. You can realize $G$ as a generalized Schottky group, acting on the real hyperbolic space $H^n$ for some $n$ via a geometrically finite action. You first take a geometrically finite kleinian group $G_0$ whose parabolic subgroups are exactly $\mathbb{Z}^{d_1}$,...,$\mathbb{Z}^{d_n}$. Now for each subgroup $\mathbb{Z}^{d_i}$, you choose some large $k_i$ and consider powers $e_1^{k_i},...,e_{d_i}^{k_i}$ of the standard generators $e_1$,...,$e_{d_i}$. If those $k_i$ are chosen sufficiently large, then an easy application of the pingpong lemma shows that the subgroup $G_1$ generated by these elements is a free product. Since the subgroup of $\mathbb{Z}^{d_i}$ generated by $e_1^{k_i}$,...,$e_{d_i}^{k_i}$ is still isomorphic to $\mathbb{Z}^{d_i}$, $G_1$ is isomorphic to $G$.
Note that you can play the same game with torsionfree nilpotent groups, replacing the real hyperbolic plane by some simply connected Riemannian manifold of pinch negative curvature, but there are some differences. To simplify, assume that $G$ is of the form $G=G_1*\mathbb{Z}$, where $G_1$ is torsionfree nilpotent. Then, $G_1$ can be realized as the stabilizer of a cusp in a finite volume manifold of pinched negative curvature. This is a consequence of several results, including those of P. Ontaneda's spectacular paper Pinched smooth hyperbolization, see this question on mathoverflow for more details. You then choose a loxodromic element $g$ and applies the pingpong lemma to $g$ and a subgroup of $G_1$ generated by elements that are far from the identity. You thus get a free product of the form $\mathbb{Z}*G_1'$. However, the main difference is that it is not clear to me if you can always have $G_1'$ isomorphic to $G_1$. To my knowledge, different lattices in the same nilpotent Lie group can be nonisomorphic, see for instance there.