Yes, that is correct. The projection $\pi : \mathbf R^n\to T^n$ is a covering map. A necessary and sufficient condition for a differential form on $\mathbf R^n$ to descend to $T^n$ is that it be constant on the fibres of $\pi$.
(Of course, for a general map it doesn't make sense to say that a differential form is constant on the fibres, but for a Galois covering, the tangent spaces at two different points in a given fibre can be identified using the covering group.)