What is an example showing the failure of the functoriality of the cone construction? The main problem with triangulated categories is the fact that the cone construction is not-universal, only homotopy universal. What is an explicit example of two non-equal induced morphisms of triangles
$$
\begin{matrix}
A \xrightarrow{f} & B \to & C(f) \xrightarrow{[+1]} & A[+1] \\
\downarrow &\downarrow & \downarrow\downarrow &\downarrow \\
X \xrightarrow{g} & Y \to & C(g) \xrightarrow{[+1]} & X[+1]
\end{matrix}
$$
and what is the homotopy?
 A: Example: Consider the following diagram over the derived/homotopy category of a field ${\mathbb k}$:
$$\begin{matrix} \Sigma^0{\mathbb k} & \xrightarrow{\quad} & 0 & \xrightarrow{\quad} & \Sigma^1{\mathbb k} & \xrightarrow{\ \ 1\ \ } & \Sigma^1{\mathbb k}\\ \downarrow & (\ast) & \downarrow 
 & &\scriptsize{0}\downarrow\downarrow \scriptsize{1} && \downarrow \\ 0 & \to & \Sigma^1{\mathbb k} & \xrightarrow{\ \ 1\ \ } & \Sigma^1{\mathbb k} & \xrightarrow{0} & 0\end{matrix}$$
(Here $\Sigma^n\ {\mathbb k}$ means the stalk complex over ${\mathbb k}$ concrentrated in cohomological degree $-n$).
Both rows are distinguished triangles, and the diagram commutes regardless which of two non-vertical vertical arrows at third position you choose. 
Conceptual explanation (representation theoretic):
In order to define a canonical cone for a commutative square living in a homotopy category, like $(\ast)$ above, you need to first represent it as a strictly commutative square of chain complexes. Once that choice has been made, you can convince yourself that the cone construction is functorial.
Now, what does e.g. lifting $(\ast)$ to a strictly commutative square of chain complexes mean? A strictly commutative square of chain complexes over ${\mathbb k}$ is the same as a chain complex $X$ of representations of the commutative square $\square := \scriptsize\begin{matrix}\bullet & \to & \bullet \\ \downarrow && \downarrow \\ \bullet & \to & \bullet\end{matrix}$ over ${\mathbb k}$. The condition that the underlying commutative square over $\textsf{D}({\mathbb k})$ of $X$ is isomorphic to $(\ast)$ means that $H^0(X)$ and $H^{-1}(X)$ are the simple representations of $\square$ at the upper left resp. lower right corner. 
Hence, the question comes down to the following:

Underlying problem: Is any complex over the square $\square$ determined by its cohomology?

As you might know, the answer is no, since the algebras for which that is the case are the hereditary ones, those where $\text{Ext}^n\equiv 0$ for $n>1$. The commutative square, however, is not hereditary. In fact, the example diagram corresponds to a non-zero class in $\text{Ext}^2_\square$ between the simple representations at upper left / lower right corner - you can work that out yourself or check the details in Example I.7.3.5 on p. 135 here.
It's also possible to understand why cones exist, yet only up to non-canonical isomorphism:

Underlying problem: Is any complex over $\bullet\to\bullet$ determined by its cohomology?

Here, the answer is yes, but the uniqueness isomorphism is not canonical!

Punchline: The non-uniqueness in defining the colimit of an $I$-shaped diagram in a homotopy category relates to the homological complexity of $I$. While $I=\bullet\to\bullet\ $  is hereditary, allowing for definition of cones up to non-unique isomorphism, already $I=\square$ is non-hereditary, resulting in non-functoriality of the cone.

