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.