Let $(\mathcal{D},T)$ be an additive category with an autoequivalence $T:\mathcal{D}\to\mathcal{D}$. One could theoretically modify the (TR2) axiom to instead require that $X\xrightarrow{u}Y\xrightarrow{v}Z\xrightarrow{w}TX$ is distinguished if and only if $ Y\xrightarrow{v}Z\xrightarrow{w}TX\xrightarrow{Tu}TZ$ is distinguished, thereby deleting the minus sign. As far as I can tell, up to deletion of certain minus signs in some formulas, nothing would change in the theory and existing results of triangulated categories. Nonetheless, if later we would like to equip $K(\mathcal{A})$—the homotopy category of cochain complexes with terms in an abelian category $\mathcal{A}$—with the structure of a triangulated category, we would then need to define $T=-[1]$, i.e., $T^n=(-1)^n[n]$, because of the way distinguished triangles work in $K(\mathcal{A})$. The (TR2) axiom written in the standard way (with the minus sign) allows to spell $T^n=[n]$ for the rotation functor in $K(\mathcal{A})$.
In one sentence: the minus sign in TR2 is because we want our notation to be comfortable with the primary examples of triangulated categories, $K(\mathcal{A})$ and $D(\mathcal{A})=K(\mathcal{A})[\text{qis}^{-1}]$. One could ask now: well, then why do we define $d_{X^\bullet[n]}=(-1)^nd_{X^{\bullet+n}}$? Without the $(-1)^n$ we could have (TR2) without the minus sign. I answer this question here. Briefly put, we need the minus sign in the formula for $d_{X^\bullet[n]}$ because this is what allows one to define the d.t.'s in $K(\mathcal{A})$ (hence in $D(\mathcal{A})$). Note that, so far, we did set up two definitions: (i) $d_{X^\bullet[n]}$ with some minus sign and (ii) the d.t.'s as specified in the linked answer. It turns out that as a consequence of these choices, a triangle $X\xrightarrow{u}Y\xrightarrow{v}Z\xrightarrow{w}X[1]$ in $K(\mathcal{A})$ is distinguished if and only if $ Y\xrightarrow{v}Z\xrightarrow{w}X[1]\xrightarrow{-u[1]}Z[1]$ is distinguished (as it is shown in the second paragraph of the proof of 014S).