Let $\mathcal{C}$ be a stable $\infty$-category. Then $\mathcal{C}$ has a homotopy category $h \mathcal{C}$, which is triangulated. The collection of morphisms $f: X \rightarrow Y$ of $\mathcal{C}$ can be organized into an $\infty$-category $\mathrm{Fun}( \Delta^1, \mathcal{C} )$. The operation $f \mapsto \mathrm{Cone}(f)$ can be obtained from a functor of $\infty$-categories $\mathrm{Fun}( \Delta^1,\mathcal{C} ) \rightarrow \mathcal{C}$. You can pass to homotopy to get a functor of ordinary categories
$$\mathrm{Cone}: h \mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow h \mathcal{C}.$$

The functor $\infty$-category $\mathrm{Fun}( \Delta^1, \mathcal{C} )$ is equipped with an evaluation functor $e: \Delta^1 \times \mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow \mathcal{C}$. You can take homotopy categories here, to get a functor of ordinary categories $[1] \times h\mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow h\mathcal{C}$, which can be identified with a functor (again of ordinary categories)
$$ U: h \mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow \mathrm{Fun}( [1], h\mathcal{C} ).$$
Here $[1]$ denotes the category $\{ 0 < 1 \}$ consisting of two objects and one morphism between them.

The phenomenon you're asking about is due to the fact that $U$ is not an equivalence of categories. Moreover, the functor $\mathrm{Cone}$ does not factor through $U$. If $f: X \rightarrow Y$ and $f': X' \rightarrow Y'$ are morphisms of $\mathcal{C}$, then morphisms from $f$ to $f'$ on in the $\infty$-category
$\mathrm{Fun}( \Delta^1, \mathcal{C} )$ can be thought of triples $(u,v,h)$, where $u: X \rightarrow X'$ and $v: Y \rightarrow Y'$ are morphism of $\mathcal{C}$ and $h$ is a homotopy from $f' \circ u$ to $v \circ f$. In these terms, the functor $U$ is given on morphisms by the construction $[(u,v,h)] \mapsto ( [u], [v] )$
where $[s]$ denotes the homotopy class of a morphism $s$. In particular $U$ "forgets" the data of the homotopy $h$, and fails to be a faithful functor.