How can I produce a summation function from this set of production rules for a grammar?
OK, so the question now has a set of recurrence relations.  To solve these, you can plug them in to each other until you get a single recurrence relation.  First, substitute $c(n)=b(n-4)$ to get rid of all of the $c$'s:
$$a(n) = 2 b(n-4) + 3 b(n-8)$$
$$b(n) = a(n-2) + b(n-8).$$

Now substitute in to get rid of the $a$'s:

$$b(n) = 2 b(n-6) + 3 b(n-10) + b(n-8).$$

So, you've gotten down to a single recurrence relation for the $b$'s.  Solve this using standard methods; see Solving or approximating recurrence relations for sequences of numbers for details.  Then, once you have a solution for the $b$'s, you can reverse the substitutions and obtain a solution for the $a$'s, $c$'s, and finally $s$'s.

If you want to find an exact equation for the $b$'s (rather than an asymptotic estimate), the method of generating functions will probably be useful.  You'll have to find the roots of a 10th-degree polynomial ($x^{10} - 2 x^4 - x^2 - 3$, I think), and the solution will likely be an expression of the form $c_1 r_1^n + \dots + c_{10} r_{10}^n$ where $r_1,\dots,r_{10}$ are the roots and $c_1,\dots,c_{10}$ are the appropriate constants.  This might get a bit messy, but it should be doable.

Or: You can ask Wolfram Alpha to solve the recurrence relation for you.  Sometimes it is able.  Unfortunately, for this one it seems to have failed: output.