Let me try to reinstate honor to the solution that proposed the basis change, by **reducing** the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the **continuous version**. (as in Peter Humphries' solution)

Let $S$ be the subspace of vectors $(x_1,\ldots,x_k)$ such that
$$a_1x_1+\dots+a_kx_k=0$$
for all integer vectors $(a_1,\ldots,a_k)$ for which
$$a_1\theta_1+\dots+a_k\theta_k=0.$$
The integer points in $S$ form a lattice $L$, for which we choose some
generating
basis vectors $b_1,\dots,b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$. Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $(\theta_1,\dots,\theta_k)\in S$ with respect to this basis.
Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals, and by the generic continuous Kronecker-Weyl Theorem,
$(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$.
Transforming back, $(t(\theta_1,\dots,\theta_k))$ is uniformly distributed modulo $L$ in the fundamental region
$$ F = \{\,\lambda_1b_1+\dots+\lambda_rb_r \mid
0\le\lambda_i<1\,\}$$
of the lattice. Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have an integer point inside $F$), but
"opposite" boundary points _are_ mapped to the same point because they differ by a basis vector $b_i$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.