Why not just prove the result from scratch? It's only a couple of pages and involves only basic Fourier analysis for locally compact abelian groups.
Let
$$\mathbb{T}^n = \left\{(z_1,\ldots,z_n) \in \mathbb{C}^n : |z_l| = 1 \text{ for all $1 \leq l \leq n$}\right\}$$
be the $n$-torus. Let $t_1, \ldots, t_n$ be arbitrary real numbers, and let $H$ be the topological closure in $\mathbb{T}^n$ of the subgroup
$$\widetilde{H} = \left\{\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \in \mathbb{T}^n : y \in \mathbb{R}\right\}.$$
The Kronecker--Weyl theorem states that
(1) $H$ is a closed connected subgroup of $\mathbb{T}^n$, namely an $r$-dimensional subtorus of $\mathbb{T}^n$, where $0 \leq r \leq n$ is the dimension over $\mathbb{Q}$ of the span of $t_1, \ldots, t_n$, and that
(2) for any continuous function $h : \mathbb{T}^n \to \mathbb{C}$, we have that
$$\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{h\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \: dy} = \int_{H}{h(z) \: d\mu_H(z)},$$
where $\mu_H$ is the normalised Haar measure on $H$.
To prove this, we begin by observing that $H$ is a closed connected subgroup of $\mathbb{T}^n$ as it is the topological closure of $\widetilde{H}$, which is a subgroup of the compact abelian group $\mathbb{T}^n$, being the image of the continuous group homomorphism $\phi : \mathbb{R} \to \mathbb{T}^n$ given by $\phi(y) = \left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right)$.
Next we recall that a character $\chi : \mathbb{T}^n \to \mathbb{T}$ is of the form
$$\chi(z_1,\ldots,z_n) = z_1^{k_1} \cdots z_n^{k_n}$$
for some $(k_1,\ldots,k_n) \in \mathbb{Z}^n$. Conversely, for any $(k_1,\ldots,k_n) \in \mathbb{Z}^n$, the function $\chi : \mathbb{T}^n \to \mathbb{T}$ defines a character of $\mathbb{T}^n$. In particular, the dual group of $\mathbb{T}^n$ is isomorphic to $\mathbb{Z}^n$, and hence every character $\chi : \mathbb{Z}^n \to \mathbb{T}$ is of the form
$$\chi(k_1,\ldots,k_n) = z_1^{k_1} \cdots z_n^{k_n}$$
for some $(z_1,\ldots,z_n) \in \mathbb{T}^n$.
We claim that the annihilator $H^{\perp}$ of $H$ (namely the set of characters of $\mathbb{T}^n$ that are trivial on $H$) is isomorphic to $\left\{k \in \mathbb{Z}^n : t_1 k_1 + \cdots + t_1 k_n = 0\right\}$, and consequently $H$ is isomorphic to a torus $\mathbb{T}^r$ for some $0 \leq r \leq n$. Indeed, each character $\chi \in H^{\perp}$ is of the form $\chi(z_1,\ldots,z_n) = z_1^{k_1} \cdots z_n^{k_n}$ for some $(k_1,\ldots,k_n) \in \mathbb{Z}^n$ with the property that for all $y \in \mathbb{R}$,
$$1 = \chi\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) = e^{2\pi i (t_1 k_1 + \cdots + t_n k_n) y},$$
and hence $t_1 k_1 + \cdots + t_n k_n = 0$. Conversely, if $t_1 k_1 + \cdots + t_n k_n = 0$, then the homomorphism $\chi(z_1,\ldots,z_n) = z_1^{k_1} \cdots z_n^{k_n}$ satisfies $\chi \vert_{H} = 1$. Now by construction, $H^{\perp}$ is isomorphic to $V \cap \mathbb{Z}^n$ for some vector subspace $V$ of $\mathbb{Q}^n$ of dimension $n - r$, where $r$ is the dimension over $\mathbb{Q}$ of the span of $t_1, \ldots, t_n$, and so $H^{\perp} \cong \mathbb{Z}^{n-r}$. Consequently, $\widehat{H} \cong \mathbb{Z}^n / H^{\perp} \cong \mathbb{Z}^r$, as the dual group of $\mathbb{T}^n$ is isomorphic to $\mathbb{Z}^n$, and hence $H \cong \mathbb{T}^r$, thereby proving (1).
For (2), we require the Fourier transform $\widehat{h} : \mathbb{Z}^n \to \mathbb{C}$ of a continuous function $h : \mathbb{T}^n \to \mathbb{C}$, defined by
$$\widehat{h}(\chi) = \int_{\mathbb{T}^n}{h(z) \overline{\chi(z)} \: dz}.$$
The Poisson summation formula for a closed subgroup of $\mathbb{T}^n$ states that
$$\int_{H}{h(z) \: d\mu_H(z)} = \int_{H^{\perp}}{\widehat{h}(\chi) \: d\mu_{H^{\perp}}(\chi)},$$
where $\mu_H$ is the normalised Haar measure on $H$ and $\mu_{H^{\perp}}$ is the induced Haar measure on $H^{\perp}$, which is the counting measure as $H^{\perp}$ is discrete. Now if $h : \mathbb{T}^n \to \mathbb{C}$ is a trigonometric polynomial, which is to say a function of the form
$$h(z) = \sum_{k \in \mathbb{Z}^n} c_k z_1^{k_1} \cdots z_n^{k_n}$$
for $z = (z_1, \ldots, z_n) \in \mathbb{T}^n$, where all but finitely many of the coefficients $c_k \in \mathbb{C}$ are zero, we claim that
$$\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{h\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \, dy} = \int_{H}{h(z) \: d\mu_H(z)},$$
where $\mu_H$ is the normalised Haar measure on $H$. From this, we may easily obtain the result in the general case where $h$ is merely a continuous function by the density of the trigonometric polynomials in the space of continuous complex-valued functions on $\mathbb{T}^n$ with regards to the supremum norm. This yields (2).
To prove the claim, we let $\chi : \mathbb{T}^n \to \mathbb{T}$ be a character corresponding to $\widetilde{k} \in \mathbb{Z}^n$. Then
$$\widehat{h}(\chi) = \int_{\mathbb{T}^n}{h(z) \overline{\chi(z)} \, dz} = \int_{\mathbb{T}} \hspace{-0.1cm} \cdots \hspace{-0.1cm} \int_{\mathbb{T}}{\sum_{k \in \mathbb{Z}^n} c_k z_1^{k_1} \cdots z_n^{k_n} \overline{z_1^{\widetilde{k_1}} \cdots z_n^{\widetilde{k_n}}} \, dz_1 \cdots dz_n}.$$
We may interchange the order of summation and integration, as there are only finitely many nonzero members in this sum, and evaluate this integral in order to find that $\widehat{h}(\chi) = c_{\widetilde{k}}$. Recalling that $H^{\perp}$ is isomorphic to $\left\{k \in \mathbb{Z}^n : t_1 k_1 + \cdots + t_n k_n \in \mathbb{Z}\right\}$, so that the Haar measure $\mu_{H^{\perp}}$ on $H^{\perp}$ is simply the counting measure, we therefore obtain by the Poisson summation formula that
$$\int_{H}{h(z) \: d\mu_H(z)} = \sum_{\substack{k \in \mathbb{Z}^n \\ t_1 k_1 + \cdots + t_n k_n = 0}} c_k.$$
On the other hand,
\begin{align*}
\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{h\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \: dy} & = \lim_{Y \to \infty} \frac{1}{Y} \sum_{k \in \mathbb{Z}^n} c_k \int^{Y}_{0}{e^{2\pi i (t_1 k_1 + \cdots + t_n k_n) y} \: dy} \\
& = \sum_{\substack{k \in \mathbb{Z}^n \\ t_1 k_1 + \cdots + t_n k_n = 0}} c_k
\end{align*}
as required, where we justify the interchanging of order of summation and integration by noting that there being only finitely many nonzero members in this sum.
Best Answer
It seems that you are after this result which can be found, for example, as Theorem 3.12 in Gareth A. Jones, Josephine M. Jones: Elementary Number Theory, Springer-Verlag, London, 1998. Springer Undergraduate Mathematics Series. (It is in the section 3.5 entitled An extension of the Chinese Remainder Theorem.)