A short exact sequence that cannot be made into an exact triangle. (Weibel 10.1.2) The following exercise is in Weibel Chapter 10.
Regard the groups $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ as cochain complexes in degree 0. Show that the short exact sequence 
$$
0 \rightarrow \mathbb{Z}/2\mathbb{Z} \stackrel 2 \rightarrow \mathbb{Z}/4\mathbb{Z} \stackrel 1 \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow 0
$$
cannot be made into an exact triangle in the homotopy category of chain complexes.
My intuition for this is that the map $\text{cone}(f) \rightarrow A[-1]$ has to be non-trivial either cohomologically or on the level of chains, but I can't seem to come up with an obstruction for the above sequence.
Any advice? Also, why / how do maps from the mapping cone to $A[-1]$ differ from "regular maps"? Why are they special?
 A: I read this recently in the LMS notes on Triangulated Categories. Unfortunately, I'm going to assume the next section of Weibel where we see that the homotopy category $K$ of an abelian category $A$, is a triangulated category.
Assume that we could extend this exact sequence to a distinguished triangle
$$\mathbb{Z}/2\longrightarrow \mathbb{Z}/4\longrightarrow \mathbb{Z}/2\longrightarrow \mathbb{Z}/2[-1].$$
The morphism $w:\mathbb{Z}/2\rightarrow\mathbb{Z}/2[-1]$ obviously has the be the zero map in the category of chain complexes, so $w=0$ in our homotopy category $K$. Example 10.1.5 in Weibel tells us that $\mathbb{Z}/2\rightarrow\mathbb{Z}/2\rightarrow 0\rightarrow \mathbb{Z}/2[-1]$ is a distinguished triangle, and in fact the other triangle in Example 10.1.5 is an example of a general fact in triangulated categories:
If a map in a distinguished triangle is trivial, then one of the other maps is a split monomorphism and the other a split epimorphism.
This is sort of the key ingredient, so let's look at the following commutative diagram of distinguished triangles.
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{ccccccc}
\mathbb{Z}/2 & \ra{2} & \mathbb{Z}/4 & \ra{} & \mathbb{Z}/2 & \ra{0} & \mathbb{Z}/2[-1]\\
\da{id} & &  & & \da{} & & \da{id} \\
\mathbb{Z}/2 & \ra{id} & \mathbb{Z}/2 & \ra{} & 0 & \ra{} & \mathbb{Z}/2[-1]\\
\end{array}
$$
Once we know that $K$ is a triangulated category, we know that there exists a map $s:\mathbb{Z}/4\rightarrow \mathbb{Z}/2$ such that $s\circ 2=id$ (this is TR3 in Weibel). The existence of such a map tells us that $\mathbb{Z}/2\oplus\mathbb{Z}/2\cong\mathbb{Z}/4$ in $K$, so these two chain complexes must be homotopic as chain complexes (concentrated in degree 0). Since both complexes are concentrated in a single degree, there is no possibility for a homotopy equivalence between them. This means that we must have already had $\mathbb{Z}/2\oplus\mathbb{Z}/2\cong\mathbb{Z}/4$ as an isomorphism of chain complexes. This is our contradiction.
This problem will happen anytime we have an exact sequence of chain complexes which is not split exact. Note that this is not a problem once we move from the homotopy category $K$ to the derived category $D$. In the derived category, distinguished triangles give us SES's of chain complexes, and visa versa.
As I'm currently reading this triangulated categories book, I'm not sure what to say about your other questions. I'm still trying to understand the big picture. 
