When existence of loxodromic, WPD elements implies an action is acylindrical Definitions
Say that $(X,d)$ is a $\delta$-hyperbolic space and that $G$ is a finitely generated group acting on $X$ by isometries.
Recall that an action of $G$ on $X$ is called acylindrical if the following holds:


*

*For every $\epsilon>0$, there exists $R,N>0$ such that for every $x,y\in X$ satisfying $d(x,y) \geq R$, there are at most $N$ elements $g\in G$ such that $d(x,gx) \leq \epsilon$ and $d(y,gy) \leq \epsilon$.


Also recall that an element $h\in G$ is called a WPD element if the following holds:


*

*For every $\epsilon >0$ and every $x\in X$, there is an $M\in \mathbb{N}$ such that $|\{ g\in G:d(x,gx)<\epsilon, d(h^Mx,gh^Mx)<\epsilon\}|<\infty$.


Via Osin's paper Acylindrically hyperbolic groups, we know that the following are equivalent definitions of being acylindrically hyperbolic:


*

*$G$ admits a non-elementary acylindrical action on a hyperbolic space $X$.

*$G$ is not virtually cyclic and admits an action on a hyperbolic space $Y$ such that at least one element of $G$ is loxodromic and satisfies the WPD condition.


Questions
It is clear to me that if $G$ acts acylindrically and non-elementarily on $X$, that 2 follows, with $X=Y$. My first question is the following:
Question 1: Given 2, the hyperbolic space $X$ in 1 need not be the same hyperbolic space $Y$ from 2. Is it known how one can construct $X$ given $Y$?
My second question is related.
Question 2: If we know that 2 holds, is it known when we can take $X=Y$? That is, if we know that 2 holds, what is required to be able to say that the action of $G$ on $Y$ is acylindrical?
Motivation
In the setting of the mapping class group, we can take both $X$ from 1 and $Y$ from 2 to be the curve complex, (with some mild assumptions on the complexity of the surface). The fact that the action is acylindrical is due to Bowditch in the paper Tight geodesics in the curve complex, and the fact that pseudo-Anosovs act as WPD elements on the curve complex is due to Bestvina-Fujiwara in Bounded cohomology of subgroups of mapping class groups.
I am interested in various subcomplexes of the curve complex and subgroups of the mapping class group acting on these subcomplexes and would like to know if these actions are also acylindrical. 
 A: To answer question 1, there is a description of the construction of $X$ in Section 5 of Osin's paper, which I recalled in this question. Basically, $X$ is obtained coning-off both a maximal virtually cyclic group $H$ containing your WPD element and a group "transverse" to $H$ in some sense.
Taking the example of the free group $G=F(a,b)$ of two generators and considering $a$ to be WPD, then Osin's contructions yields the Cayley graph $Cay(G, \langle a\rangle, \langle b\rangle)$. You see here that $X$ is not quite $Y$ (the group $G$ itself).
I'm not sure how to understand question 2, but if you're asking if given a WPD element, you can deduce the action is acylindrical, this is not true. If you're asking if there are general methods to prove that in particular cases, I don't know any (except using definitions of acylindrical actions of course).
For your motivation, you can indeed take the same space $X=Y$ to be the curve complex, but this is because the action is acylindrical (and 1 implies 2 as you say). To prove that the action of a subgroup on a subcomplex you're interested in is acylindrical, I would suggest without more details to try to reproduce Bowditch's proof. 
A: Question 2 is vague, but here is an example which, I think, shows that it is not reasonnable to expect positive results in this direction without (strong) additional assumptions.
Consider the infinite free product $G:= \underset{n \geq 2}{\ast} \left( \mathbb{Z} \oplus \mathbb{Z}_n \right)$. You can think of $G$ as the fundamental group of the following graph of groups:

Let $T$ denote the associated Bass-Serre tree on which $G$ acts. Because vertex-stabilisers are finite, every infinite-order element of $G$ is WPD. However, the action of $G$ on $T$ is not acylindrical (because, for every $n \geq 2$, the axis of a generator of the $\mathbb{Z}$ in $\mathbb{Z} \oplus \mathbb{Z}_n$ is fixed by the $\mathbb{Z}_n$). Even worse, $G$ does not admit an acylindrical action on a hyperbolic space for which all its infinite-order elements are WPD (otherwise, we would have a bound on the index of $\langle g \rangle$ in the centraliser $C(g)$ independ on the infinite-order element $g$), so every hyperbolic space on which $G$ would act acylindrically must be quite different from $T$.
The key point is that $G$ admits a universal action (i.e., an action on a hyperbolic space for which every generalised loxodromic element is loxodromic) but it does not admit a universal acylindrical action (i.e., an acylindrical action on a hyperbolic space for which every generalised loxodromic element is loxodromic). There exist finitely generated such example, such as Dunwoody's inaccessible group.
