We define $\sigma_i(w): F\rightarrow \mathbb{Z}$ the number of appearances of the generator $a_i$ in $w$ when $a_i , a_i^{-1}$ cancels each other and we work with reduces word (means there are no consecutive appearances of $aa^{-1}$, or $a^{-1}a$ in $w\in F$.
So, for the question: In case $w\in [F,F]$, $w=w_1w_2w_1^{-1}w_2^{-1}$ then its pretty simple to demonstrate that for each $i$, $\sigma_i(w_1w_2w_1^{-1}w_2^{-1})=0$.
I am having some trouble with the second direction - I was trying to apply induction over the number of generators appears in $w\in F$ which satisfies $\sigma_i(w)=0$ for all $1\le i\le n$.
Induction base: for $w$ which is generated by an $a_i$ because we work with reduced words, one learn the $w=e$ (the empty word) and indeed $e\in [F,F]$.
Now for the induction step - $w$ is generated by $k$ generators. So we may decompose $w$ to: $w=w_1a_iw_2a_i^{-1}w_3$ or $w=w_1a_i^{-1}w_2a_iw_3$ when $w_2$ does not include any appearance of $a_i$, we may find such a decomposition becasue we work with reduces words. Let us assume that $w = w_1a_iw_2a_i^{-1}w_3$ so if we multiply by $w_2^{-1}w_2$ we get: $w=w_1a_iw_2a_i^{-1}w_2^{-1}w_2w_3$ when $a_iw_2a_i^{-1}w_2^{-1}$ is a commutator! Now one may repeat this method for $w_1$, $w_3$ for each $a_i,a_i^{-1}$ pair, and get $w = u_1[c_1]u_2[c_2]...u_k[c_k]u_{k+1}$ when $a_i$ does not appear in any of the $u_i$s and all the [c_i]s are commutator. That is the point where I'm stuck, I wish I could switch the order such that all the commutators will be set left (or right) to the $u_i$s and the apply the induction step, but I don't know whether it's possible to do so.
Any other Idea would be appreciated!
You don't need induction here. The only thing you need is to remember that $[F,F]\trianglelefteq F$ and the quotient group $F/[F,F]$ is abelian. First of all let's write $S=\{a_1,...,a_n\}$. Now let $w\in F$ such that $\sigma_i(w)=0$ for all $i$. We can write $w=b_1b_2...b_k$ when each $b_j$ is an element from $S$ or an inverse of an element of $S$. Now we have:
$w[F,F]=b_1...b_k[F,F]=b_1[F,F]...b_k[F,F]$
But because the quotient group is abelian you can change the order of the cosets and bring it to the form $a_1^{r_1}[F,F]a_2^{r_2}[F,F]...a_n^{r_n}[F,F]=a_1^{r_1}a_2^{r_2}...a_n^{r_n}[F,F]$. But as you know $\sigma_i(w)=0$ for all $i$, and that means $r_1=r_2=...=r_n=0$. Hence:
$w[F,F]=a_1^{r_1}a_2^{r_2}...a_n^{r_n}[F,F]=[F,F]$
And that means $w\in[F,F]$.