Long words represent by multiplication of short words

Question

Give a free group $G$ and one of its subgroup $H$ satisfies

1. $rank(G)=n$
2. $[G:H]=k$

Fix a generators of $G$ so we can talk about the length of elements in $G$.Then do there exist constants $A,B,C$ which depends only on $n,k$ and the choice of generators. Satisfying:
$\forall h\in H$ whose length is larger that A, We can always find elements $h_1,\dots,h_s \in H$ with length less than $B$ s.t. the following holds:

1. $h=h_1h_2\dots h_s$
2. $|h|>|h_1|+|h_2|+\dots+|h_s|-C$

Answer 1

I don't think this is true. Let $G$ be free on $x,y$, and $H$ the subgroup of index $2$ generated by $x,y^2,y^{-1}xy$.

Consider $h = y^{-1}x^ny \in H$ for some large $n$. The putative word $h_1h_2\cdots h_s$ would at length at most $C$ more than $|h| = n+2$, and would freely reduce to $y^{-1}x^ny$, so it would have to be the word $y^{-1}x^ny$ with at most $C/2$ cancelling generator pairs inserted. Since the word $y^{-1}x^ny$ has no prefixes that lie in $H$, the word $h_1h_2\cdots h_s$ would have at most $C/2$ prefixes in $H$ giving $s \le C/2$, which is not possible if the $|h_i|$ are bounded.