How is the DTFT of a periodic, sampled signal linked to the DFT?
I think that your misunderstanding comes from the fact that you assume that $f[n]$ is periodic. If $f[n]$ were periodic then its DTFT would be a Dirac comb. What is happening here is that you want to find the $N$-periodic sequence whose DFT coefficients equal equidistant samples of the DTFT of $f[n]$. And it turns out that that sequence is given by
$$\tilde{f}[n]=\sum_{k=-\infty}^{\infty}f[n-kN]\tag{1}$$
which is - as expected - an aliased version of the original sequence. Remember that sampling in one domain corresponds to aliasing in the other domain.
A derivation of $(1)$ is given in this answer.