Here is an elementary approach, which shows how to find the nature of $nf_n(t)$ as $n\to\infty$. But I'm not going to bound error terms or such so this remains an outline until those details are filled in.
Write $a(k) = \mu^k(1-t\mu^k)^n$ and consider $k$ as a real variable. $a(k)$ is easily seen to have a unique maximum at $k=k_0$ where $\mu^{-k_0}=1/((t(n+1))$. Now calculate $$a(k_0) = \frac{(n/(n+1))^n}{t(n+1)} \sim \frac{e^{-1}}{t(n+1)}$$ $$a(k_0+u) = a(k_0) \mu^u\left(\frac{1-\mu^u/(n+1)}{1-1/(n+1)}\right)^n \sim a(k_0) \mu^u \exp(1-\mu^u)),$$ the limits being for $n\to\infty$ with $u$ not too wild.
So now we have (modulo checking of error terms), $$nf_n(t) \sim \frac{1}{t} \sum_u \mu^u e^{-\mu^u},$$ where the sum is over $u\ge -k_0$ such that $k_0+u$ is an integer. The restriction $u\ge -k_0$ should be far enough in the tail that it doesn't matter, so we have $$nf_n(t) \sim \frac{1}{t} F(y),$$$$nf_n(t) \sim \frac{1}{t} F_\mu(y),$$ where $y$ is the fractional part of $k_0$ and $$F(y) = \sum_{j=-\infty}^\infty \mu^{y+j}e^{-\mu^{y+j}}.$$$$F_\mu(y) = \sum_{j=-\infty}^\infty \mu^{y+j}e^{-\mu^{y+j}}.$$ (Doesn't this last sum have a name? I'm sure I've seen it before.)