The part that it is a continuation is an error from when I mentioned the first claim elsewhere in the first draft, and the second claim was the first claim back then. It is also very poorly phrased in retrospect (in quite a lot of ways); it is actually connected to the subsequence $$$t'$$$ in the second claim. Here's what I meant when I wrote that (paraphrasing from my notes):

Let's say we have $$$st, sa \in L$$$. Consider the $$$t'$$$ from the second claim, and the positions of the subsequence that was shifted by 1. If we move $$$a$$$ in that subsequence while preserving the order of the other elements, then the resulting $$$t"$$$ will satisfy $$$st" \in L$$$. (Looking at it in another way, we are shifting one part of the remaining subsequence to before $$$a$$$, while keeping the order of the remaining elements the same, which is essentially what the claim says).

If we consider the previous example, where $$$s = pqr$$$, $$$t = uvwxyz$$$ and $$$t' = avwuyx$$$, we can move around the sequence $$$a$$$ within the characters $$$a, u, x$$$, while preserving the relative position of $$$u$$$ and $$$x$$$, any number of times to get $$$t"$$$ such that $$$st"$$$ is in $$$L$$$. Thus $$$t"$$$ can be $$$uvwayx$$$, $$$uvwxya$$$ and of course $$$t'$$$.

Simply map a $$$w \in L$$$ to the set of its characters, call this mapping $$$f$$$. We claim that $$$f$$$ is the required mapping to go from $$$L$$$ to $$$F$$$. $$$|s| > |t|$$$ in $$$L$$$ implies, since $$$L$$$ is a simple language, that $$$f(s) \setminus f(t)$$$ is the set of characters that are in $$$s$$$ but not in $$$t$$$, and there can be at most one occurrence of any character in $$$S$$$ in the list of characters in $$$s$$$ but not in $$$t$$$ (with multiplicity). So for every $$$tx$$$ for $$$x \in s$$$ and $$$x \not \in t$$$, we have a unique element $$$f(t) \cup \{x\}$$$ in $$$F$$$. Now for any $$$A, B$$$ in $$$F$$$ with $$$|A| > |B|$$$, we can use any preimage of $$$A$$$ and any preimage of $$$B$$$ in $$$L$$$ to be able to claim that there is a $$$x$$$ in $$$A \setminus B$$$ such that $$$A \cup \{x\} \in F$$$. Abuse of notation: we will also denote the map from $$$L$$$ to the set of sets as $$$f$$$.

For the other direction, considered an unordered greedoid $$$(S, F)$$$, and the maximal simple hereditary language $$$L$$$ such that $$$f(L) \subseteq F$$$. It is easy to check that $$$L$$$ is an ordered greedoid. Let's call this mapping $$$g$$$.

Note that $$$f$$$ and $$$g$$$ applied in this order will map $$$L$$$ to itself — this can be shown by induction on word size.

Since this gives an explicit natural bijection, we can use either definition for a greedoid. Most sources (such as Wikipedia) use an unordered version, but I find the ordered version to be much more intuitive for greedy algorithms.