I believe I have a "moral solution". The final paragraph must be made rigorous (or perhaps there is a cleaner way to finish things off), but it looks like this is the right direction:

**Reduction:**

Fix $p\in (0,1/2)$. For large $n$, it is not hard to see that $$f(1,n)>f((p-4p^2) n,n)(1-3p).$$

Indeed, consider the balls an urn's setup of Fedor Petrov. For the first $pn$ steps, we have at most $pn$ white balls and at least $(1-p)n$ total balls, so we expect to eat at most $\frac{p}{1-p}pn<(2-\epsilon)p^2n$ white balls total (where $\epsilon>0$ since $p<1/2$). By a Chernoff bound, a.a.s. we should eat at most $2p^2n$ balls in the first $pn$ steps, and (assuming we run all these steps), we end up creating at least $(p-2p^2)n$ white balls. 

Meanwhile, the probability that the original white ball is eaten in the first $pn$ steps is at most $(pn)\frac{1}{(1-p)n}\le 2p$. For large $n$, the a.a.s. event fails with probability at most $p$, thus a union bound gives that with probability $\ge 1-3p$, that at time $pn$ we have $\ge (p-4p^2)n$ white balls and $\le n$ black balls.

**Finishing touches (hand-wavy):**

It remains to show that for fixed $c>0$ that $f(cn,n)=1-o(1)$. I will sketch an approach which I believe could made rigorous.

Let's renormalize this as $f(pn,qn)$ where $p+q= 1$. If we run this for $t =\epsilon n$ steps, then another Chernoff bound says we should almost surely have $\approx ((1-\epsilon)p+\epsilon q)n$ white balls and $\approx (1-\epsilon)q n$ black balls (up to $O(\epsilon^2)$ errors). But if one looks at the function
$$\frac{(1-x-100x^2)p+(x-100x^2)q}{(1-x+100x^2)q}-\frac{p}{q}$$for any $p,q\in (0,1)$ satisfying $p+q=1$ (on [Desmos][1], say), it is clear that that $x=0$ is not a local maximum. 

Thus for small $\epsilon$, we should get that $f(pn,qn)\ge f(p'n',q'n')-o(1)$ for $p'>p$ and $n'\ge n/2$. And so presumably as we iterate things (replacing $p'$ by $p''$ and so on), we'll get that $f(pn,qn)\ge f((1-\delta)n^*,\delta n^*)-o(1)$ where $n^* = \Omega_{p,\delta}(n)$. From here, (I assume) standard balls and urns or some first moment argument should give that $f((1-\delta)n^*,\delta n^*) \le c_\delta +o(1)$ where $c_\delta \to 0$ as $\delta \to 0$. 
  


  [1]: https://www.desmos.com/calculator/bh2f2o5cs1