The following answer is incomplete.
Clearly, $0\le f\le1$. Let $h:=1-f$, so that $0\le h\le1$, $h(a,0)=0$, and $h(0,b)=1$ if $b\ge1$. Here and in what follows, $a$ and $b$ are nonnegative integers.
We want to show that $h(1,n)\to0$ (as $n\to\infty$). The idea is to construct a suitable majorant $g$ of $h$.
Let $$g(a,b):=\frac{2 (1+a) b}{2 (1+a) b+a (1+a+b \ln b)}$$ if $b>0$, with $g(a,0):=0$. It is enough to show that $$h(a,b)\le g(a,b) \tag{1}\label{1}$$ for all $a,b$.
Let us do this by induction on $m:=m(a,b):=a+2b$. If $m=0$, then \eqref{1} is trivial. Also, \eqref{1} turns into an equality if $a=0$ or $b=0$. If now $a\ge1$ and $b\ge1$, then \eqref{1} follows by induction from the given recurrence for $f$ (and the same recurrence for $h$) and the inequality $$g(a,b)\ge\frac a{a+b}\, g(a-1,b)+ \frac b{a+b}\,g(a+1,b-1) \tag{2}\label{2} $$ for $a,b\ge1$. (To use the induction, note that $m(a-1,b)=m(a+1,b-1)=a+2b-1<m(a,b)$.)
Remark: Inequality \eqref{2} actually fails to hold, by a small amount of $\asymp10^{-13}$ for some very large $a,b$. I am leaving this answer for now, hoping that someone can modify $g$ a bit to make \eqref{2} hold.
