It's easier to prove this result using a different definition of $\sigma$, so let's first prove the two definitions are equivalent. I claim that the definition of $\sigma$ given above satisfies
$$
\sigma(E) = 3 m(\overline{E})
$$
for all measurable $E \subseteq S^2$, where $m$ is $3$-dimensional Lebesgue measure and $\overline{E} := \{ tx : 0 \leq t \leq 1, x \in E \}$. To prove this, evaluate the right hand side using spherical coordinates:
$$
m(\overline{E}) = \int_{\mathbb{R}^3} 1_E = \int_{0}^{\infty} \int_0^\pi \int_0^{2\pi} 1_\overline{E}(r\sin\phi\cos\theta, r\sin\phi\sin\theta, r\cos\phi) r^2 \sin\phi \,d\theta \,d\phi \,dr.
$$
By definition, $(r\sin\phi\cos\theta, r\sin\phi\sin\theta, r\cos\phi) \in \overline{E}$ if and only if $(\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi) \in E$ and $0 \leq r \leq 1$. Therefore the above is equal to
$$
\int_0^1 r^2 \int_0^\pi \int_0^{2\pi} 1_E(\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi) \sin\phi \,d\theta \,d\phi \,dr = \sigma(E) \cdot \int_0^1 r^2\,dr = \frac13 \sigma(E)
$$
as claimed.
Now, in order to prove that $\widehat{\sigma}$ is radial, we need to prove that $\widehat{\sigma}(A\xi) = \widehat{\sigma}(\xi)$ for any $3\times3$ orthogonal matrix $A$. First we have
$$
\widehat{\sigma}(A\xi) = \int_{S^2} e^{-2\pi i \langle x, A\xi \rangle} \,d\sigma(x) = \int_{S^2} e^{-2\pi i \langle Ax, \xi \rangle} \,d\sigma(x)
$$
because orthogonal matrices move across inner products. Another property of orthogonal matrices is that they preserve distances between points, and therefore they also preserve the $3$-dimensional Lebesgue measure (in the sense that $m(A^{-1} E) = m(E)$ for any measurable $E$). Therefore since we've shown that $\sigma(E)$ is just a constant multiple of $m(\overline{E})$, $\sigma$ must also be preserved by the action of orthogonal matrices, which allows us to conclude
$$
\int_{S^2} e^{-2\pi i \langle Ax, \xi \rangle} \,d\sigma(x) = \int_{S^2} e^{-2\pi i \langle x, \xi \rangle} \,d\sigma(x) = \widehat{\sigma}(\xi)
$$
as desired.
