Restriction and extension
Considering a smooth compact hyper-surface $\mathcal{S}$ in $\mathbb{R}^d$ with surface measure $d\sigma$. Given $f \in L^1(\mathbb{R}^d)$, the Fourier transform defined as follow: $$ \begin{equation} \hat{f}(x) = \int_{\mathbb{R}^d}e^{-2\pi i x \xi}f(x)dx \end{equation} $$ which by Riemann-Lebesgue is a bounded, continuous function vanishing at infinity.
Since $\hat{f}$ is continuous on $\mathbb{R}^d$, by the Rimann-Lesbegue lemma its restriction to the compact hyper-surface $S \subset \mathbb{R}^d$ is is well-defined pointwise. Specifically, the restriction $\hat{f}\mid_{S}: S \rightarrow \mathbb{C}$ is the continuous function given by $$ \begin{equation} \hat{f}\mid_{S}(\sigma) = \hat{f}(\sigma) = \int_{\mathbb{R}^d}e^{-2\pi i x \xi}f(x)dx \end{equation} $$ for each $\sigma \in S$. This is bounded (as $\hat{f}$ is bounded) and can be integrated against the surface measure $d\sigma$ on $S$.
Thus when we restrict $\hat{f}$ to $S$, we get a meaningful function which has finite $L^q$-norm for every $q$ .
When starting with $f \in L^2(\mathbb{R}^d)$, the Fourier transform $\hat{f}$ is not well-defined point-wise in general, so there is no meaningful way to restrict an arbitrary $L^2$ function to a set of measure zero such as the hyper-surface $S$.
For especially, for any given $f \in L^2(\mathbb{R}^d)$, the Fourier transform is defined in the $L^2$ sense via the Plancherel theorem: $$ \begin{equation} \mathcal{F}: L^2(\mathbb{R}^d) \to L^2(\mathbb{R}^d), \quad | \hat{f} | _{L^2} = | f | _{L^2} \end{equation} $$ It is an isometry. So: $$ \begin{equation} \hat{f} \in L^2(\mathbb{R}^d) \end{equation} $$ Since $\hat{f}$ is only an $L^2$ function — it is not necessarily continuous, and not even bounded, and need not have a pointwise value almost everywhere.
So the expression: $$ \begin{equation} \hat{f}|_S(\sigma) = \hat{f}(\sigma), \quad \sigma \in S \end{equation} $$ does not make sense pointwise for arbitrary $f \in L^2$.
The question arises: what happens for $1 < p < 2$?
For which $p$ and $q$ do we have: $$ \begin{equation} ||\hat{f}|| _{L^q(S, d\sigma)} \lesssim ||f|| _{L^p(\mathbb{R}^d)}, \quad \forall f. \end{equation} $$
This is restriction of Fourier transforms to hyper-surfaces problem in Harmonic analysis.