Reduction of decison SIS
The assumption about gcd tell you that $2\alpha +1$ cannot belong to an ideal of $\mathbb{Z}/q\mathbb{Z}$. For example,  $2\alpha +1$ could divide $q$ and then $(2\alpha+1)t$ is never going to look like a uniformly random vector in $\mathbb{Z}_q^n$. If you want to fix ideas, assume $3$ divides $q$ and $\alpha =1$: then $(2\alpha+1)t$ always has its entries a multiple of $3$, which is absolutely not what a uniform vector would look like. The added contribution of $Ar$ may not be enough to "blur" the entries to a more uniformly random vector: it is likely that you could observe too many instances where $(2\alpha+1)t+Ar$ has entries a multiple of $3$.

In general, having this assumption make the proof easier (it might even not be possible to have a general proof without it) and relax the care on $\alpha$. It is also not very restrictive.