The key fact here is the (surprising, initially, but well known) (power) decay of $\widehat{\sigma}(\xi)$. If $u\in C^{\infty}(S)$, we can extend to a function $u_0\in C^{\infty}_0(\mathbb R^d)$, and then $$ \widehat{u\, d\sigma}= \widehat{u}*\widehat{\sigma} $$ still decays. See here for the general version of the convolution theorem needed here.

We can then extend this to arbitrary $u\in L^1$ by the argument from the traditional Riemann-Lebesgue lemma: given $\epsilon>0$, pick a $v\in C^{\infty}(S)$ with $\|u-v\|_1<\epsilon$. Since $|\widehat{u\, d\sigma}(\xi) -\widehat{v\, d\sigma}(\xi)|<\epsilon$ and $\widehat{v\, d\sigma}\to 0$, we also have $|\widehat{u\, d\sigma}(\xi)|<2\epsilon$ for all large $\xi$.

The key fact here is the (surprising, initially, but well known) (power) decay of $\widehat{\sigma}(\xi)$. If $u\in C^{\infty}(S)$, we can extend to a function $u_0\in C^{\infty}_0(\mathbb R^d)$, and then $$ \widehat{u\,d\sigma}=\widehat{u_0\,d\sigma}=\widehat{u_0}*\widehat{\sigma} $$ still decays. See here for the general version of the convolution theorem needed here.

We can then extend this to arbitrary $u\in L^1$ by the argument from the traditional Riemann-Lebesgue lemma: given $\epsilon>0$, pick a $v\in C^{\infty}(S)$ with $\|u-v\|_1<\epsilon$. Since $|\widehat{u\, d\sigma}(\xi) -\widehat{v\, d\sigma}(\xi)|<\epsilon$ and $\widehat{v\, d\sigma}\to 0$, we also have $|\widehat{u\, d\sigma}(\xi)|<2\epsilon$ for all large $\xi$.