= Solution
Let $F(t)=\sum_{k\in\mathbb Z}|f(t+2\pi k)|^2$. By the <Tonelli theorem>, $F$ is integrable on $[0,2\pi]$, since its integral is $\int_{\mathbb R}|f|^2<\infty$. The <Plancherel theorem> and periodization give
$$
\langle\phi(\cdot-n),\phi(\cdot-m)\rangle
=\frac1{2\pi}\int_{\mathbb R}|f(t)|^2e^{i(m-n)t}\,dt
=\frac1{2\pi}\int_0^{2\pi}F(t)e^{i(m-n)t}\,dt.
$$
Consequently, if $F=1$ almost everywhere these <inner products> are $\delta_{mn}$, proving <orthonormality>. Conversely <orthonormality> says that every <Fourier coefficient> of $F-1$ is zero. The <uniqueness of Fourier coefficients in L1> implies $F-1=0$ almost everywhere. Hence
$$
\boxed{\{\phi(\cdot-n)\}_{n\in\mathbb Z}\text{ is orthonormal}\quad\Longleftrightarrow\quad\sum_k|f(t+2\pi k)|^2=1\text{ a.e.}}
$$
This proves <orthonormal translates and Fourier periodization>, including both directions and the normalization constant associated with the printed <Fourier transform>.
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