> 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.

This is conjecture based on observing a pattern, but it feels so natural that it inspires some confidence even based on a relatively small sample.

Define the sequence of words $r_i = \textrm{reverse}(w_i)$. Initialise $i = 0$, $s = \textrm{reverse}(W[1..n])$ is the $n$th splitter. While $s$ is non-empty, if $r_i$ is a prefix of $s$ emit $A$, remove $r_i$ from the start of $s$, and increment $i$; otherwise emit $B$, remove $r_{i+1}$ from the start of $s$, and increment $i$ twice.

Worked example: the 26th splitter is $10010010100101001001010010$. It doesn't start with $r_0 = 0$, so we emit $B$ and skip $r_1 = 10$ leaving $010010100101001001010010$. This does start with $r_2 = 010$, so we emit $A$ and reduce $s$ to $010100101001001010010$. This doesn't start with $r_3 = 10010$, so we emit $B$ and skip $r_4 = 01010010$ leaving $1001001010010$. This does start with $r_5 = 1001001010010$, so we emit $A$ and the empty string remains. The Wythoff composite for the 26th splitter is $BABA$.

For practical implementation, it may be noted that $r_i$'s first symbol is $i \bmod 2$ and the lengths, obviously, follow the Fibonacci sequence.