Solution:

The value $12$ is fixed by the function, since $12$ has $6$ positive divisors and therefore $f_1(12)=12$.

First find all $n\le50$ with $f_1(n)=12$, meaning $n$ has $6$ divisors. These are $$12,18,20,28,32,44,45,50.$$

Now check whether $f_1(n)$ can be one of these values before reaching $12$. Since $f_1(n)$ is twice a divisor count, the only useful possibilities in that list are $18$ and $20$, meaning $n$ has $9$ or $10$ divisors.

For $n\le50$, the additional possibilities are $36$, which has $9$ divisors, and $48$, which has $10$ divisors. Therefore there are $8+2=10$ values of $n$.

Thus, D is the correct answer.