This is either trivial or wrong:
Let $f:\mathbb S^1 \to \mathbb R$, where $\mathbb S^1 = [-1/2,1/2]/(-1/2\sim1/2)$ is the one-dimensional sphere, be such that $f\in L^1\big(\mathbb S^1\big)$; does it make sense to define $\tilde f:\mathbb R \to \mathbb R$ such that $\tilde f(x)= f(x)$ if $x\in (-1/2,1/2]$, and $\tilde f(x)= 0$ otherwise, and to say that for any $\kappa\in \mathbb Z$ $$\mathcal F \tilde f(u)|_{u=\kappa}:=\int_{\mathbb R}\tilde f(x)e^{2\pi i \kappa x }dx=\int_{\mathbb S^1} f(x) e^{2\pi i \kappa x }dx=:\mathcal F f(\kappa)?$$
I have some results about Fourier transforms of functions on $\mathbb R$ that I now need for functions on $\mathbb S^1$ and this argument would allow me to avoid reproving them. The definition of $\tilde f$ and the equality above seem to make sense computationally but for some reason my intuition tells me that they do not mathematically.
The manipulation you described is correct, but it usually does not give an easy way to obtain results for the Fourier transform on $S^1$ from the results for the Fourier transform on $\mathbb{R}$. For example, $\tilde f$ is typically discontinuous even if $f$ is continuous, which is an obstacle to relating the smoothness of $f$ to the decay of its Fourier coefficients.
What happens in such cases is: $\mathcal F\tilde f$ has slow decay at infinity, but dips toward zero at integer points, thus aligning with the quickly decaying Fourier coefficients $\hat f(n)$. A simple illustration is $f\equiv 1$, where all Fourier coefficients except one are zero. Since $\tilde f=\chi_{[-1/2,1/2]}$, its Fourier transform is the sinc function with some normalization: it is pretty heavy-tailed (not in $L^1(\mathbb{R})$) but duly vanishes at the integers.
A relevant result on the relation between the Fourier transform of a function and the Fourier series of its periodization is the Poisson summation formula.