= Solution
Use the <smooth periodic partition of unity> $\Psi$ above and define the <Fourier coefficients> of the <periodic distribution> by
$$
c_n=\frac1{2\pi}\langle u,\Psi(x)e^{-inx}\rangle,\qquad n\in\mathbb Z.
$$
These are the <smooth-window Fourier coefficients of a periodic distribution> for period $2\pi$. The finite <Schwartz seminorm> bound for $u$, restricted to the fixed <compact support> of $\Psi$, gives $|c_n|\leq C(1+|n|)^N$ for some $C,N$. Indeed, only finitely many <derivatives> of $\Psi e^{-inx}$ occur, and each contributes at most a fixed power of $|n|$.
For a <Schwartz function> $\phi$, the preceding <Fourier series> formula and $\widehat{\widehat\phi}(n)=2\pi\phi(-n)$ yield
$$
P\widehat\phi(x)=\sum_{n\in\mathbb Z}\phi(-n)e^{inx}.
$$
All <derivatives> of this series converge uniformly. The localization formula in the previous answer therefore gives
$$
\begin{aligned}
\langle\widehat u,\phi\rangle
&=\langle u,\widehat\phi\rangle
=\langle u,\Psi P\widehat\phi\rangle\\
&=\sum_n\phi(-n)\langle u,\Psi e^{inx}\rangle
=2\pi\sum_n c_n\phi(n).
\end{aligned}
$$
Consequently the <Fourier transform of a periodic distribution> is
$$
\boxed{\widehat u=2\pi\sum_{n\in\mathbb Z}c_n\delta_n\quad\text{in }\mathcal S'(\mathbb R).}
$$
The polynomial bound on $c_n$ and rapid decay of $\phi(n)$ make every displayed pairing absolutely convergent. In particular there are no <derivatives> of <Dirac delta distributions> in this formula.
The <inverse Fourier transform> of $\delta_n$ is $(2\pi)^{-1}e^{inx}$. Applying continuous <Fourier inversion> on <tempered distributions> to the symmetric partial sums gives
$$
\boxed{u=\sum_{n\in\mathbb Z}c_ne^{inx}\quad\text{in }\mathcal S'(\mathbb R).}
$$
Explicitly, for every <Schwartz function> $\theta$, the pairing of the partial sum with $\theta$ is $\sum_{|n|\leq M}c_n\widehat\theta(-n)$, which converges absolutely to $\langle u,\theta\rangle$. The <Fourier transform> formula also proves that $c_n$ is independent of the chosen <smooth periodic partition of unity>. For a locally integrable $2\pi$ <periodic function> it reduces, by splitting the integral into periods, to $c_n=(2\pi)^{-1}\int_0^{2\pi}u(x)e^{-inx}\,dx$.
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