Reasonable question. The three parts of the proof are not done consecutively; they are mutually recursive. Part 2 assumes that the induction hypothesis already holds for F. This IH could have been proved by any of the parts, including part 3. Part 2 then extends the claim from F to ¬F.
For example, consider the formula ¬(p /\ ¬q) for propositional variables p and q. Let P(F) denotes the induction statement, i.e., for every initial segment W of F we have o(W) >= c(W). Then the proof proceeds as follows (bottom up).
1. P(p) base case
2. P(q) base case
3. P(¬q) induction step from 2
4. P((p /\ ¬q)) induction step from 1 and 3
5. P(¬(p /\ ¬q)) induction step from 4.
Viewed top down, it can be represented as a tree.
So, part 2 uses IH that could have been proved by part 3, and part 3 uses two IHs that could have been proved by part 2, etc.
P(¬(p ∧ ¬q))
P((p ∧ ¬q))