Finding the number of elements of order two in the symmetric group $S_4$

Hint: you're forgetting to include and count those permutations that are the product of two disjoint two-cycles in $S_4$:

$$(1\, 2)(3 \,4),\; (1\, 3)(2 \, 4),\; (1\, 4)(2 \,3)\; \in S_4,$$ which comprise three additional elements in $S_4$, each of order $2$, as are the six 2-cycles of order $2$ you counted.

Recall that the order of a permutation which is the product of disjoint cycles is equal to the $\;\operatorname{lcm}\;$ (i.e., the least common multiple) of the orders of its cycles.

As any permutation can be written as a product of disjoint cycles, and the product of disjoint cycles has order equal to the least common multiple of those cycles' lengths (=orders), the group $\,S_n\,$ has elements of order $\,r\,$ iff we can find disjoint cycles in it with their lengths' l.c.m. equal to $\,r\,$.

Thus, for example, in $\,S_4\,$ the order of the element with largest order is $\,4\,$ (as distinct cycle decomposition like $(a, b, c, d)$) , whereas in $\,S_5\,$ we have elements of order $\,6\,$ , say $\,(12)(345)\,$ , etc.

What is left is to count the number of cycles in $\,S_n\,$ ...