(Note: This does not answer your question. Edit: It still doesn't answer your question, but I am attempting to make my notation less horribly broken, so future readers don't get too confused.)
I'm not entirely sure which parts of the diagram you are considering, so I'll try to fix some notation. We start with maps $X \to Y \to Z$, and my understanding of the octahedral axiom is that by taking cones, we get maps $X/Z \to Y/Z \to X/Y \to X/Z[1]$$Y/X \to Z/X \to Z/Y \to Y/X[1]$ that form a distinguished triangle. The objects can be arranged into an octahedral diagram, with alternating triangles that commute or are distinguished, but I don't know which object is the upper vertex.
Since the objects $X/Y$$Y/X$, $X/Z$$Z/X$, and $Y/Z$$Z/Y$ together with the maps between them are only defined up to nonunique isomorphism, I don't see why one should get a map of octahedra from the given data. I have heard that one needs an extra "octahedron-gluing" axiom (or some underlying stable infinity or model category) to pass from a commuting diagram $(X \to Y \to Z) \to (X' \to Y' \to Z')$ to a map of octahedra. However, I don't recall a precise obstruction to getting I think the problem is that the "commuting map of triangles" axiom produces two maps $Z/X \to Z'/X'$ from the commuting squares like the one containing $X/Z \to Y/Z$$(X \to Z) \to (X' \to Z')$ and $X'/Z' \to Y'/Z'$$(Z/Y[-1] \to Y/X) \to (Z'/Y'[-1] \to Y'/X')$, and there is no reason for those two maps to coincide.
You might be able to construct a counterexample using information on page 6 in Toën's paper, where there is a list of problems with localization.