Maybe my answer is beside the point, because the question speaks about the "limit distribution of $t(\theta_1,\dots,\theta_d)$" (for real $t$?). Shouldn't it be $n(\theta_1,\dots,\theta_d)$ for $n=1,2,\ldots$? See https://mathworld.wolfram.com/Kronecker-WeylTheorem.html. The long answer "from scratch" by Peter Humphries proves a different theorem that seems to be more in line with real parameters $t$. Also, the sequence $n(\theta_1,\dots,\theta_d)$ can fill a disconnected manifold, which wouldn't properly be called a "subtorus".
If the question is indeed about the sequence $n(\theta_1,\dots,\theta_d)$, the book by Kuipers and Niederreiter (Uniform Distribution of Sequences, 1974), contains the generic version of what seems to be the Kronecker-Weyl Theorem as Example 6.1 on p. 48. The proper condition is the real numbers $1,\theta_1,\dots,\theta_d$ are linearly independent over the rationals.
The notes on p.51 mention that
a discussion of the exceptional case in this example was also carried out by Weyl,
referring to his classical paper: Hermann Weyl. Über die Gleichverteilung von Zahlen mod. Eins. Mathematische Annalen, 77:313–352, 1916.
Indeed, §5 of that paper, "Die Ausnahmefälle" (the exceptional cases), contains a Theorem 18 (pp. 340-341). It deals with the more general case where each coordinate is not just a linear function of $n$ but an arbitrary polynomial. The conclusion is that the points cover a finite number of affine-linear $r$-dimensional manifolds, (possibly with different integer multiplicities), all these manifolds are parallel, and each of them is filled with uniform density. The theorem specifies how to determine $r$ and the multiplicities.
Here is a statement of this theorem (with different notation), specialized and reformulated for the case of an arithmetic progression $n(\theta_1,\dots,\theta_d)$ as opposed to arbitrary polynomials. The multiplicities are then not necessary.
Let $\vec\theta=(\theta_1,\dots,\theta_d)$. Let $C$ be the set of vectors $\vec x\in\mathbb R^d$ such that $$\langle \vec a,\vec x\rangle\equiv b \pmod 1$$ for all integer vectors $\vec a\in\mathbb Z^d$ and rational numbers $b$ for which the equation $$\langle \vec a,\vec\theta\rangle= b$$ holds. Then all numbers $b$ appearing in these equations have a least common denominator $g$. The sequence $n\vec\theta$ is uniformly distributed modulo 1 in the disjoint union of parallel subspaces $C\cup 2C\cup \dots \cup gC$.
The denominator $g$ is the smallest number $g\ge1$ for which $gC$ contains an integer point (or equivalently, for which $gC$ modulo 1 contains the origin and is therefore equivalent to its corresponding linear subspace $C-C$). If rational dependencies exist only among the numbers $\theta_1,\dots,\theta_d$ and not with the number 1, then the right-hand side $b$ is always $0$, and we set $g=1$.
It is clear that $n\vec\theta$ cycles through the sets $C,2C,3C,...$. Thus it suffices to look at the generating vector $g\vec\theta$ and prove that its multiples are uniformly distributed modulo one in $gC=C-C$. It is an exercise to reduce this case to the independent case, along the lines of the reduction that I gave for the continuous version. (In fact many of the arguments in that prove appear in Weyl's proof.)