Recall that $\mathscr S(\mathbb R) \subset L^2(\mathbb R)$. Assume the continous and bounded functions $f: \mathbb R \to \mathbb C, \,f \in C(\mathbb R), f(t) \not= 0$ only if $a \leq t \leq b$ and that $|f(t)| \leq 1$ for all $t$.
We define for all $x,y \in L^2(\mathbb R)$
\begin{equation} Wf(t,\eta) = \int \limits_{\mathbb{R}} \! f(t + \frac{\tau}{2}) \overline{ f(t - \frac{\tau}{2}) } e^{-i2\pi \eta \cdot \tau} d\tau. \end{equation}
We have a nonlinear mapping
\begin{equation} (f \mapsto Wf) : L^2(\mathbb R) \to L^2(\mathbb R \times \mathbb R), \end{equation}
where we extend the $L^2$ spaces without rigour onto the space of tempered distributions:
\begin{equation} (f \mapsto Wf) : \mathscr S'(\mathbb R \to \mathscr S'(\mathbb R \times \mathbb R). \end{equation}
Based on Vobo's answer. You need the theorem 5.1 by Treves Francois to prove the extension from the $L^2$ spaces onto the space of tempered distributions:
How can you prove the extension by using Vobo's mentioned things?
This is a rather technical argument, let me give you a basic idea:
Theorem: Let $X$ and $Y$ be complete metric spaces, $D\subseteq X$ a dense subset, $f:D\to Y$ uniformly continuous. Then $f$ has a unique continuous extension to $X$, i.e. there is a unique continuous $g:X\to Y$ with $g(d)=f(d)$ for all $d\in D$.
The proof strategy for this is quite obvious: For $x\in X$ let $(d_n)_n$ be a sequence in $D$ converging to $x$. The uniformly continuous $f$ transforms the Cauchy sequence $(d_n)_n$ to a Cauchy sequence $(f(d_n))_n$. Now set $g(x):=\lim_n f(d_n)$ which exists and prove that this limit does not depend on the special sequence $(d_n)_n$.
In your case, let $X=\mathcal{S}'(\mathbb{R}), Y=\mathcal{S}'(\mathbb{R}\times\mathbb{R})$ and $D=L^2(\mathbb{R})$. Next step would be to prove the uniform continuity of $W$.
Unfortunately, $\mathcal{S'}$ is not a metric space. But luckily, the above theorem extends to complete topological vector spaces like $\mathcal{S}'$ with the same proof strategy. But talking about completeness or uniform continuity in non-metric spaces requires concepts like uniform spaces and this is - I would assume - beyond your scope.
So the main question is to show that your map $W$ is uniformly continuous with respect to the strong topology on $\mathcal{S}'$. This means:
Given a $0$-neighbourhood $U$ in $\mathcal{S}'(\mathbb{R}^2)$, i.e. an $\varepsilon>0$, a bounded set $B_2\subset\mathcal{S}(\mathbb{R}^2)$ - that is some $C_{\alpha,\beta}>0$ such that for all $\varphi\in B_2$ you have $\sup_x |x^\alpha D^\beta \varphi(x)| \leq C_{\alpha,\beta}$ - there is a $0$-neighbourhood $V$ in $\mathcal{S}'(\mathbb{R})$ - that is some $\delta>0$ and a bounded set $B_1\subset\mathcal{S}(\mathbb{R})$ - such that $$ f,g\in L^2(\mathbb{R})\,\text{ with }\,\forall\psi\in B_1: |\int_\mathbb{R} (f-g)(t) \psi(t) dt|<\delta \,\,[\text{i.e.}\, f-g\in V] \\ \implies \forall \varphi\in B_2: |\int_{\mathbb{R}^2} (Wf(t,\eta)-Wg(t,\eta))\varphi(t,\eta) dtd\eta |<\varepsilon\,\,[\text{i.e.}\, Wf-Wg\in U]. $$
I have the feeling that this is true, but I am not able to prove this - sorry.