A more-or-less obvious upper bound for $f(n)$ is $3n-2$: divide the cake into $n$ pieces of size $\frac1{n+1}$ plus $n-1$ pieces of size $\frac1{n(n+1)}$ and plus $(n-1)$ pieces of size $\frac1{(n+1)n(n-1)}$. So, the question is if this upper bound $3n-2$ is exact. — Taras Banakh May 4 at 6:30

@TarasBanakh I'll just mention that there was a comment in chat related to this question (and more specifically to your original $3n+2$ estimate). Just in case you would like to respond there. — Martin Sleziak 51 mins ago

@MartinSleziak Thank you for pointing me this comment in chat. Indeed, Cuize Han was right that there was a mistake in my argument, but not the upper bound $3n-2$: the idea was to divide the case into $(n+1)$ equal parts (this requires $n+1$ cuts), then take one part and divide it into $n$ equal pieces of size $\frac1{n(n+1)}$ which requires $n-1$ cuta and then take another piece of size $\frac1{n+1}$ and divide it into $n-1$ equal parts or size $\frac1{n(n-1}$, which requires $n-2$ cuts. So, together, it will be exactly $n+1+n-1+n-2=3n-2$ cuts. — Taras Banakh 11 mins ago

On the other hand, the same upper bound $3n-2$ can be obtained by superposing regular polygons with $n-1$, $n$ and $n+1$ vertices, sharing a common vertex, as was suggested y Fedor Petrov in his comment. — Taras Banakh 5 mins ago