How to think about the octohedral axiom for triangulated categories? I am currently learning about triangulated categories and came across the following diagram in Gelfand & Manin:

Here the triangles with a star inside are distinguished while the triangles with the curved arrow inside are just commutative. Gelfand & Manin then claims that the octohedral axiom can be stated as: Any diagram of the form (upper cap) can be extended to an octohedral diagram, where this includes some other commutative requirements. 
He then goes on to claim that a diagram of the form (upper) cap can be thought of as a morphism of triangles where the second vertical arrow is the identity on $Y$. I am having trouble seeing how or why this is true.
Say we have the diagram (upper cap) above. This gives us the two triangles
$$
X \stackrel{u}{\longrightarrow} Y \stackrel{v}{\longrightarrow} Z' \stackrel{w}{\longrightarrow} \Sigma X
$$
and
$$
Y \stackrel{u'}{\longrightarrow} Z \stackrel{v'}{\longrightarrow} X' \stackrel{w'}{\longrightarrow} \Sigma Y
$$
Now we can rotate the second triangle to get a distinguished triangle
$$
\Sigma^{-1} X' \stackrel{-\Sigma^{-1}w'}{\longrightarrow} Y \stackrel{u}{\longrightarrow} Z \stackrel{v'}{\longrightarrow} X'
$$
I am assuming these are the two triangles that we can get the morphism between with the second arrow being the identity on $Y$. But what are the morphisms here? I'm not able to see that. 
 A: I fought with motivation for the octahedral axiom for a long time, but then I saw this explanation.
Suppose you have a map of spaces $f : X\to Y$; a simple and yet remarkable fact is that every such map can be replaced, up to a homotopy equivalence, by a cofibration. More precisely, you can define the mapping cylinder $M_f$ of $f$ and the mapping cone of $f$ by $C_f = M_f / X$ (this is well-defined as $X$ embeds in the cylinder of $f$): somehow, this reflects the fact that when you are interested in homotopy theory of spaces, you can take any map to be a certain kind of well-behaved inclusion, by replacing the codomain $Y$ by an homotopy equivalent space, and the map  $f$ by an homotopic one.
Once you have done this, you can now define a sequence
$$X \to Y \overset{i_f}\to C_f\to C_{i_f}\to\dots$$
and the universal property of the quotients you performed entails that, given $X \overset{f}\to Y\overset{g}\to Z$, there exists a canonical map $\varphi : C_f \to C_{gf}$. Now,

Octahedral axiom: $C_g \simeq C_\varphi$.

In spaces, this is something you can prove by hand. When you move to chain complexes, the mapping cone construction makes sense also for co/chain maps. If $X \overset{f}\to Y\overset{g}\to Z$ are two composable co/chain maps, you can still prove that $C_g\simeq C_\varphi$ for these chain maps. That's it. 
