Whenever $P$ is a summand of $Q$, you can construct $P\oplus\Sigma P$ in one step from $Q$: if $e$ is the idempotent that projects onto $P$, then the cofiber of $1-e$ is $P\oplus\Sigma P$.
You can apply this to the summand $Z$ of $X$.
There are obstructions to the more general case when $Y$ is $0$ in $K_0$ : for any $Y$ in any stable $\infty$-category $C$, embed $C$ in a bigger one whose $K_0$ vanishes. The question of whether $Y\in \langle Y\oplus\Sigma Y\rangle$ does not depend on whether you are in $C$ or this bigger one, but the vanishing of the class of $Y$ in $K_0$ does.
Basically, by Thomason's result it boils down to the class of $Y$ in $K_0(\langle Y \rangle)$. If it vanishes, then by Thomason's result you get what you want, and if it doesn't, that's an obstruction. The vanishing of $Y$ in $K_0(\langle Y\rangle )$ means, taking $R = End(Y)$, that the class of $R$ is zero in $K_0(Perf(R))$.
Say for a second that $R$ is connective, then this means (using the fact that we can compute this $K_0$ via projectives) that there exists a finite $n$ with $R^n\simeq R\oplus R^n$, $R$-linearly and from this it is easy to construct $R$ from $R\oplus \Sigma R$.
Here's one construction: observe that it follows that $R^n\oplus R^n\simeq R^n$, then construct $R^n\oplus\Sigma R^n$ as a sum of $R\oplus \Sigma R$'s, and then from that $R^n\oplus R^n \oplus \Sigma R^n$ , and from that together with $R^n\oplus\Sigma R^n$, you get $R^n$ as a cofiber, and then you can chop that off $R\oplus R^n$ (which you have, because it's $R^n$ !) to get $R$.
I don't know about the general case. But more or less I think that a witness for the $0$-ness of the class of $R$ will give you a construction of $R$ from $R\oplus \Sigma R$.