Translation (or shift) on chain complexes and the translated (or shifted) differential

Great question! I thought about it at some point, and I found a more down to earth explanation for myself. Maybe it is wrong, so I will be happy to see other answers!

First of all, the issue boils down to $k = 1$, since the translation by $k\in \mathbb{N}$ should be $\underbrace{[1]\cdots [1]}_k$.

Recall the construction of the cone of a morphism of complexes $f^\bullet\colon X^\bullet\to Y^\bullet$. We define it by taking objects $C^n = Y^n \oplus X^{n+1}$ and differentials $d^n = \begin{pmatrix} d^n_Y & f^{n+1} \\ 0 & -d^{n+1}_X \end{pmatrix}$. Here we already put a minus sign in front of $d^{n+1}_X$, but I don't think there's some deep reason behind it: without this sign, we simply don't get the identity $d^{n+1}\circ d^n = 0$.

Then, the cone sits in the short sequence of complexes $$0 \to Y^\bullet \rightarrowtail C (f) \twoheadrightarrow X^\bullet [1] \to 0$$ You can check that the natural projection $C (f) \twoheadrightarrow X^\bullet [1]$ is a morphism of complexes (the projections $Y^n \oplus X^{n+1} \to X^{n+1}$ commute with the differentials) only with the convention that $d^n_{X^\bullet [1]} = -d^{n+1}_X$.

It is clear that this sign convention is supposed to simplify things when one works with cones and distinguished triangles. When Verdier defines translation in his thesis, he changes the sign of the differentials, but he doesn't give an explanation.

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Weibel in his book says "this sign convention is designed to simplify notation later on" (p. 10 of his book). But then, for cohomological complexes he defines $X^\bullet [1]^n$ to be $X^{n-1}$ (probably because his book is more focused on the homological numeration?), which is not very common.