in cycle notation, we have:

r = (1 2 3 4), s = (1 4)(2 3).

so sr = (1 3), a transposition, which is clearly of order 2.

your mistake is in your calculation of (sr)^2 algebraically.

it is NOT the case that (sr)^2 = s^2r^2. typically, that is only true in abelian groups.

what you should have is this:

(sr)^2 = (sr)(sr) = s(rs)r (by associativity)

= s(sr^-1)r (from rs = sr^-1)

= (ss)(r^-1r) = (1)(1) = 1.