Such a "splitter" turns out to be simply a reversal of an initial word of $W$, so that the first few splitters are $,0,10,010,0010,10010,\ldots$. The corresponding Wythoff composites are
$$A,B,AA,AB,BA,AAA,BB,AAB,ABA,BAA,AAAA,ABB,\ldots$$
So, if someone can tell specifically how to generate this sequence, the problem will be solved.
I construct some parallel families of finite sequences in "generations", where the full sequence is obtained by concatenating all of the generations in order.
Let $W$ denote the infinite Fibonacci word.
Let $P_0 = [w_0]$, $P_1 = [w_1, w_1 w_0]$, $P_{k+2} = w_{k+2} P_k + w_{k+2} P_{k+1}$. It's easy to show by induction that the lengths of the words in $P = P_0 + P_1 + P_2 + \cdots$ are $1, 2, 3, \ldots$ and that the words themselves are all prefixes of $W$. (A useful lemma to show the latter is that $w_n w_{n-1} \cdots w_0$ is a prefix of $w_{n+2}$).
Let $r_k = \textrm{reverse}(w_k)$; equivalently $r_0 = 0$, $r_1 = 10$, $r_{k+2} = r_k r_{k+1}$. (Note that the first symbol of $r_i$ is $i \bmod 2$). Then the family $R_k$ is defined inductively as $R_0 = [r_0]$, $R_1 = [r_1, r_0 r_1]$, $R_{k+2} = R_k r_{k+2} + R_{k+1} r_{k+2}$. By construction, the $i$th word in $R_k$ is the reverse of the $i$th word in $P_k$, and therefore $R = R_0 + R_1 + R_2 + \cdots$ is the sequence of the (non-empty) splitters.
A point of interest is that the last word in $R_k$ is $r_0 r_1 \cdots r_k$, and this is a prefix of $W$ (see e.g. proposition 17 of Factorizations of the Fibonacci Infinite Word, Fici, J. Int. Seq. article 15.9.3). But it's also the reverse of a prefix, and therefore is a palindrome. Then we can work backwards along the generation and say that the $i$th last word in $R_k$ can be found at position $i$ in $W$.
The main sequence is $G_0 = [A]$, $G_1 = [B, AA]$, $G_{k+2} = G_k B + G_{k+1} A$. Clearly every (non-empty) Wythoff composite occurs exactly once in the sequence $G = G_0 + G_1 + G_2 + \cdots$. It will be useful to note that if $A$ has weight $1$ and $B$ has weight $2$ then every composite in $G_k$ has weight $k+1$. (In fact, $G_k$ contains precisely the composites of weight $k+1$ in lexicographic order of their reversals, but we won't use this).
Assuming the claim made in the question that compositions of $A$ and $B$ exactly index the splitters (which I don't know how to prove), we can show that the $i$th element of $G_k$ exactly indexes the $i$th element of $R_k$, so that $G$ is the desired sequence.
Observe that since $A$ and $B$ partition the domain of a Wythoff composite $S$, $SA$ and $SB$ partition its range. If the longest word indexed by $S$ is the splitter $s \in R_k$, then the construction of the generations of $R$ tells us that $sr_{k+1}$ and $sr_{k+2}$ are both splitters whose common prefix is precisely $s$.
Take the inductive hypothesis:
The elements of $G_k$ exactly index the corresponding elements of $R_k$ for every $0 \le k \le n$.
This can be checked for $n = 1$. Suppose it holds for a given $n \ge 1$. Then for any given splitter $s \in R_{n-1}$ with corresponding composite $S \in G_{n-1}$, the splitter $sr_n$ is indexed by $SA$, and the splitter $sr_{n+1}$ must therefore be indexed by $SBT$ for some $T$. But $T$ must be empty as otherwise there is no splitter indexed by $SB$, contradicting our assumption.
Now, for the $i$th last $s' \in R_n$ with corresponding composite $S' \in G_n$ we note that $s'r_{n+1} \in R_{n+1}$ and occurs at position $i$ in $W$. Our inductive hypothesis suffices to show that $s'$ does not occur at any position $j < i$ in $W$, since otherwise it would be indexed by multiple composites in $R_n$, requiring one to be a prefix of another, which is incompatible with the observation about their weights. Therefore $i$ is the first occurrence of both $s'$ and $s'r_{n+1}$, and $s'r_{n+1}$ must be indexed by $S'A^m$ for some $m > 0$. But in fact $m = 1$ since otherwise there is no splitter indexed by $S'A$, contradicting our assumption.
As a by-product, we obtain an effective algorithm to determine the composite which corresponds to any subword of $W$ without explicitly calculating anything more than Fibonacci numbers.
Let $s$ be the subword of interest, and initialise $i = 0$. While $s$ is non-empty, if its first symbol is $i \bmod 2$ then emit $A$, delete (up to) $f_{i+2}$ symbols from the start of $s$, and increment $i$; otherwise emit $B$, delete (up to) $f_{i+3}$ symbols from the start of $s$, and increment $i$ twice.
And as a side-note, I observe that A341258 describes a sequence of words beginning $0$, $1$, $00$, $01$, $10$, $000$, $11$, $001$, $010$, $100$, $000$, $011$, $101$, $0001$, $110$, $0010$, $0100$, $1000$, $00000$, $111$ which appears to correspond to $G$ under the substitution $0 \to A, 1 \to B$ and was submitted by Clark Kimberling in March. I assume that this is no coincidence and that you have already solved the problem by a different route.
