= Solution
Use $\widehat{f\sigma}(x)=\int_{S^1}f(\omega)e^{-2\pi ix\cdot\omega}\,d\sigma(\omega)$, with arclength measure. We first bound the transform for constant density. By rotational symmetry, $\widehat\sigma(x)=\int_0^{2\pi}e^{-2\pi ir\cos\theta}d\theta$, $r=|x|$. For $r\geq1$, remove neighborhoods of width $h=r^{-1/2}$ about the stationary points $0,\pi$. Their contribution is $O(h)$. On each remaining interval integrate by parts using the phase derivative $2\pi r\sin\theta$. The boundary terms are $O((rh)^{-1})$, and the integral of the derivative of its reciprocal is bounded by
$$
\frac Cr\int_h^{\pi-h}\frac{|\cos\theta|}{\sin^2\theta}d\theta\lesssim\frac1{rh}.
$$
Splitting at the stationary points gives the same bound on the other half of the circle. Together with the trivial estimate for bounded $r$, this proves $|\widehat\sigma(x)|\lesssim(1+|x|)^{-1/2}$.
For merely measurable $f$ there is no reason that its amplitude permits this <integration by parts>. Instead use <Gaussian positivity for even extension moments>. Set
$$
I_R(f)=\int_{\mathbb R^2}|\widehat{f\sigma}(x)|^4e^{-\pi|x|^2/R^2}dx.
$$
Expand the fourth power and integrate first in $x$. The <Fourier transform of a Gaussian> gives
$$
I_R(f)=R^2\int_{(S^1)^4} f(\omega_1)f(\omega_2)\overline{f(\omega_3)f(\omega_4)}\,e^{-\pi R^2|\omega_1+\omega_2-\omega_3-\omega_4|^2}\,d\sigma^4.
$$
All exchanges of integrals are justified by bounded amplitudes, finite surface measure and the integrable Gaussian. The kernel is nonnegative and $|f|\leq1$ almost everywhere. Taking the absolute value of the integral therefore gives $0\leq I_R(f)\leq I_R(1)$. The preceding scalar decay estimate and polar coordinates yield
$$
I_R(1)\lesssim\int_0^\infty\frac{r}{(1+r)^2}e^{-\pi r^2/R^2}dr\lesssim1+\log R\lesssim\log R\qquad(R\geq2).
$$
For the last bound split at $1$ and $R$; the tail becomes an integrable $e^{-\pi s^2}/s$ integral after $r=Rs$. Since the Gaussian is at least $e^{-\pi}$ on the radius-$R$ ball, the <local fourth-moment restriction estimate for the circle> follows:
$$
\boxed{\|\widehat{f\sigma}\|_{L^4(B(0,R))}\leq C(\log R)^{1/4}.}
$$
Every constant here is independent of $f$ and $R$.
For the requested connection to <Besicovitch sets>, let $\delta$ be small and dilate the unit segments of a bounded planar Kakeya set by $\delta^{-2}$. Their $\delta$-neighborhoods become direction-separated rectangles of length $\delta^{-2}$ and width $\delta^{-1}$. A smooth circle cap of angular width $c\delta$ has a <Fourier extension operator> of magnitude at least $c'\delta$ on such a rectangle: after removing the constant phase, the tangential phase variation is $O(c)$ and the normal variation is $O(c^2)$. A frequency modulation translates the rectangle to any desired location. Choose separated caps for the rectangle directions, and multiply their modulated densities by independent <Rademacher random variables>. Because the caps are disjoint, the input remains bounded by one.
Apply the local fourth-moment estimate in a ball of radius $C_E\delta^{-2}$ containing the rectangles, then take expectation over the <Rademacher random variables>. The fourth moment dominates the square of the sum of squared packet magnitudes, giving
$$
\delta^4\int\left(\sum_j1_{T_j}\right)^2\lesssim_E\log(1/\delta).
$$
There are $M\asymp\delta^{-1}$ rectangles, each of area $\asymp\delta^{-3}$, so the integral of their sum is $\asymp\delta^{-4}$. <Cauchy-Schwarz> forces their union area to be at least $c_E\delta^{-4}/\log(1/\delta)$. Scaling back gives $|E_{C\delta}|\gtrsim_E1/\log(1/\delta)$, the <planar Kakeya neighborhood lower bound>. The covering-number argument in Question 1 then gives both <Minkowski dimensions> equal to two.
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