In classical logic, you can derive a contradiction and then either introduce or get rid of a negation. Since you want to get a negation, you consequently might want to assume "s" and then derive a contradiction.
Anytime you have a disjunction, see if you can get both disjuncts to lead to the same conclusion, and then use disjunction elimination.
For your problem, here's an outline of a proof in one system, though I don't know what your rules say exactly, so I don't know what a proof will look like exactly in your system.
1 | s assumption (I assume s, so that I can try and find a contradiction.)
2 | (p $\lor$ q) (I'd hope I don't need the proof analysis as to how I got this)
3 || q assumption (I choose q, since if I have ¬(q∨r), I can easily get the negation of q)
4 || (q $\lor$ r) (disjunction introduction on 3)
5 | $\lnot$q (contradiction from one of the premises and 4)
6 || p assumption
7 | (p->p) (6-6 conditional introduction)
8 || q assumption (I want (q->p), so I start with q))
9 ||| $\lnot$p assumption (I want p, so I assume its negation since I already have a contradiction)
10|| p (6 and 8 contradict each other)
11| (q->p) (8-10 conditional introduction
12| p (2, 7, 11 disjunction elimination)
Now, that I have "p", I can get "r" in the same scope as "p" via the first premise (either I have modus tollens as a rule of inference, or I assume $\lnot$ r and derive a contradiction which gives me r). Then, since I have "r" I use disjunction introduction to (q $\lor$ r). But, that gives us a contradiction within the same scope as s. So, then we can introduce the negation of s within the same scope as the premises for this problem.