= Solution
Use the <Fourier transform> convention $\widehat\nu(\xi)=\int e^{-ix\cdot\xi}\,d\nu(x)$ throughout this question. The decay assumption implies square integrability, since <polar coordinates> give
$$
\int_{\mathbb R^2}|\widehat\mu(\xi)|^2\,d\xi
\lesssim\int_0^\infty\frac{r}{(1+r^{1+\epsilon})^2}\,dr<\infty.
$$
Near zero the integrand is bounded by $r$, and at infinity it is bounded by $r^{-1-2\epsilon}$. The positive $\epsilon$ is what makes the latter integrable.
By the <Plancherel theorem>, there is a function $g\in L^2(\mathbb R^2)$ whose <Fourier transform> is $\widehat\mu$. It is the density of $\mu$ with respect to <Lebesgue measure>. To justify this step rather than assume a density, for every <Schwartz function> $\varphi$, <Fourier inversion> gives
$$
\int\varphi\,d\mu
=(2\pi)^{-2}\int\widehat\varphi(-\xi)\widehat\mu(\xi)\,d\xi
=\int\varphi(x)g(x)\,dx.
$$
The <finite measure> and the locally integrable function thus define the same <tempered distribution>, so they agree as measures: $d\mu=g\,dx$. In particular $g\ge0$ almost everywhere and $\int g=\mu(\mathbb R^2)<\infty$. This is the <L2 density from a square-integrable Fourier transform> principle.
For $f\in L^\infty(\mu)$, the density of $f\,d\mu$ is $fg$. The inequality $|f|\le\|f\|_{L^\infty(\mu)}$ holds wherever $g>0$, apart from a <Lebesgue measure> zero set, so
$$
\|fg\|_2\le\|f\|_{L^\infty(\mu)}\|g\|_2.
$$
A second application of the <Plancherel theorem> yields the explicit bound
$$
\boxed{\|\widehat{f\,d\mu}\|_2
=(2\pi)\|fg\|_2
\le\|\widehat\mu\|_2\,\|f\|_{L^\infty(\mu)}.}
$$
The implicit constant in the requested estimate may depend on the measure and its Fourier-decay bound, but is independent of $f$.
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