I believe I have found a proof of the question I originally asked (see crossed out paragraph), but I have realized that what I actually need to prove is somewhat stronger. What I am actually wondering is as follows. Let $R$ be a regular language in the alphabet $\Sigma$, and $w$ be a word in $\Sigma$. For a finite automaton $M$, say that a path $(q_1,...,q_n)$ in $M$ is isolated if, for indices $i$ such that $1<i<n$, there are only two edges (one in and one out) connected to the node $q_i$, and none of the $q_i$ are initial or final states. Is it always possible to create a (nondeterministic) finite state automaton $M$ that recognizes $R$ such that all paths which output $w$ are isolated in this sense?
Edit: to be clear, the string $w$ need not be in $R$. I'm thinking principally about the case where $w$ shows up as a substring of many of the strings in $R$, and I want to "pull out" the part of the FSA which writes $w$ as it's own little branch. This would seem only to be possible when certain conditions are met, I suspect when any two instance of $w$ in a given string $r\in R$ are disjoint, but I am not sure.
Let $R$ be a regular language in the alphabet $\Sigma$, and $w$ and $v$ be words in $\Sigma$. Suppose that for all $r\in R$, whenever $w$ and $v$ are substrings of $r$ they are disjoint (they do not overlap). Is it necessarily the case that we can create some (nondeterministic) finite state automaton $M$ that recognizes $R$ such that all paths in $M$ which output $w$ are pairwise disjoint with all paths in $M$ which output $v$?
If there is a counterexample to this, it would be great to see it. If it is in fact true and is relatively trivial, just a nudge in the direction of the proof would be appreciated. If it is true and non-trivial, perhaps someone can point me to a place where it has been published?