So if I understand correctly the question you wanted to ask was:
Is it true that a triangle $$X \stackrel{u}{\to} Y \stackrel{v}{\to} Z \stackrel{w}{\to} \Sigma X$$ is split if and only if one of $u$, $v$, or $w$ is zero. The answer to this is yes.
It is clear (I think I can add details if someone wants) that if the triangle is split then one map must be zero (basically since we have an epi composing to zero). Conversely suppose that $w$ is zero, which is sufficient since we can always just rotate. Now we know by the axioms that $$Z \stackrel{-1}{\to} Z \to 0 \to \Sigma Z$$ and hence $$0 \to Z \stackrel{1}{\to} Z \to 0$$ are triangles (as an exercise check that any sequence of this form given by an isomorphism is necessarily a distinguished triangle), and $$X \stackrel{1}{\to} X \to 0 \to \Sigma X$$ is also a triangle. It is easy to check then that so is the direct sum $$ X \to X\oplus Z \to Z \stackrel{0}{\to} \Sigma X$$. The identity maps on $X$ and $Z$ then induce a map via [TR3] from this to the original triangle (since we have $w=0$ this trivially satisfies the necessary commutativity to apply [TR3]) and since two of the maps are isomorphisms so is the third. Hence any triangle as above with $w=0$ is isomorphic to one obtained by summing triangles on identity maps.
I'd recommend reading through the first chapter of Neeman's book Triangulated Categories - this is certainly covered in there as well as a bunch of other facts you might find useful in reading that paper. The reference for this result is Corollary 1.2.7 (it seemed lazy not to check since it is on my shelf) and the proof there is pretty much identical to the one here except that the facts I glossed over are proved earlier.