Lemma 1. Suppose $f$ is continuous and of moderate decrease, i.e., $|f(x)| <= {A \over (1 + x^2)}$. Then its periodization $g(x) = \sum_{k\in\Bbb Z} f(x + k)$ is continuous and periodic, and its partial sums converge uniformly to $g$.
Proof. Let $S_n(x) = \sum_{|k| \leq n} f(x + k)$. Given $n > m$, we observe the Cauchy sum of $|S_n – S_m|$ is bounded by $\sum_{|k| > n} |f(x + k)|$. But for $x \in [0, 1)$,
$$
\begin{eqnarray}
\sum_{|k| > n} |f(x + k)| &\leq& A\sum_{|k| > n} \frac{1}{1 + (x + k)^2} \\
&=& A\sum_{k > n} \frac{1}{1 + (x + k)^2} + A\sum_{k < -n} \frac{1}{1 + (x + k)^2} \\
&\leq& A\sum_{k > n} \frac{1}{1 + k^2} + A\sum_{k \leq -n} \frac{1}{1 + k^2} \\
&\leq& A\sum_{k > n} \frac{1}{k^2} + A\sum_{k \leq -n} \frac{1}{k^2} \\
&\leq& 2AT_n
\end{eqnarray}
$$
where $T_n$ is the tail sum of $\zeta(2)$. Hence $S_n$ is uniformly Cauchy on $[0, 1)$, so it converges uniformly to $g$ there, so $g$ is continuous on that interval. But $g$ is obviously periodic from its form, so the uniform convergence and continuity apply everywhere. (Possibly more details needed here.) $\blacksquare$
Lemma 2. If $|f(x+iy)| \leq \frac{A}{1 + x^2}$ for some constant $A$ on some strip $-a \leq y \leq a$, and f is holomorphic on this strip, then $|\hat f|$ decreases exponentially.
Proof. We integrate along a box from $-R \to R \to R- ai \to -R- ai \to -R$. The sides go to $0$ with $R$ and $f$ is holomorphic on the interior, so the top and bottom in the limit are equal and we have
$$
\begin{eqnarray}
\int_{-\infty}^\infty f(t)e^{-2\pi is t} dt &=& \int_{-\infty-ai}^{\infty-ai} f(t)e^{-2\pi is t} dt \\
&=& \int_{-\infty}^{\infty} f(t- ai)e^{-2\pi is (t – ai)} dt \\
&=& e^{-2\pi a s} \int_{-\infty}^{\infty} f(t- ai)e^{-2\pi is} dt
\end{eqnarray}
$$
and this is of course bounded in magnitude by $Ce^{-2\pi a s}$. $\blacksquare$
Theorem.Suppose $f$ is continuous and of moderate decrease with Fourier transform $\hat f$ and periodization $g$. Then the $n$th Fourier coefficient of $g$ is given by $\hat f(n)$.
Proof.
$$
\begin{eqnarray}
\frac{1}{2\pi} \int_{-\pi}^\pi g(t)e^{-ikt} dt &=& \frac{1}{2\pi} \int_0^{2\pi} g(t)e^{-ikt} dt \\
&=& \int_{0}^1 g(t)e^{-2\pi ikt} dt \\
&=& \int_{0}^1 \sum_{n \in \Bbb Z} f(t + n) e^{-2\pi ikt} dt \\
&=& \sum_{n \in \Bbb Z} \int_0^1 f(t + n) e^{-2\pi ikt} dt \\
&=& \sum_{n \in \Bbb Z} \int_n^{n+1} f(t) e^{-2\pi ikt} dt \\
&=& \int_{-\infty}^\infty f(t) e^{-2\pi ikt} dt \\
&=& \hat f(n).
\end{eqnarray}
$$
The interchange of integral and sum is justified by the uniform convergence, as established above. $\blacksquare$
It follows that $\sum_{n\in \Bbb Z}f(t + n) \sim \sum_{n\in \Bbb Z} \hat f(n) e^{2\pi int}$. If $f$ is decreases fast enough – e.g., if it’s Schwartz, or satisfies the conditions of Lemma 2. above – then this is actually an equality, and we can plug in $t=0$ to yield $\sum_{n\in \Bbb Z}f(n) \sim \sum_{n\in \Bbb Z} \hat f(n)$.