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.