I think you can do the subproof where you assume A and derive (A v C). You can always re-assume any premise further in the scope lines if you really want to do so, or seems convenient. So, the other subproof might look like this, with the numbering re-worked:
0 (¬B∨C) premise 1 | B assumption 2 || ¬C assumption 3 ||| (¬B∨C) assumption 4 |||| ¬B assumption 5 |||| Contradiction (not sure how you write the symbol here) 1, 4 6 |||| C assumption 7 |||| Contradiction 2, 6 8 ||| Contradiction 3, 4-5, 6-7 V elimination 9 || ¬(¬B∨C) 3-8 ¬ introduction 10 || Contradiction 0, 9 11 | ¬¬C 2-11 ¬ introduction 12 | C 11 ¬ elimination 13 C 1-12, 0, and the other subproof
Note, I'm not sure if Fitch will accept this exactly, as I've never used that program. I also feel more confident that Fitch would accept the following distinct full proof:
1 (A v B) premise 2 (¬B∨C) premise 3 | C assumption 4 | (A v C) 3 V introduction 5 | ¬B assumption 6 || A assumption 7 || (A v C) 6 V introduction 8 || B assumption 9 ||| ¬(A v C) assumption 10 ||| Contradiction 5, 8 11 || ¬ ¬(A v C) 9-10 negation introduction 12 || (A v C) 11 negation elimination 13 | (A v C) 1, 6-7, 8-12 disjunction elimination 14 (A v C) 2, 3-4, 5-13 V introduction.
Edit: I'm not sure if Fitch will accept "(A v C)", but it should come as a simple matter to drop the parentheses here. Also, if Fitch doesn't accept "(A v C)", but does accept "A v C" this actually comes as a weakness of the program, since "A v C" is not what logicians have called a "well-formed formula", while "(A v C)" does qualify as one... though your text uses the term "well-formed formula" in a slightly different way than I have here.