First, note that $$\sum_{\sigma \in S} f(\sigma)^{4}=\sum_{\sigma \in S} \sum_{1 \leq a_{1}, a_{2}, a_{3}, a_{4} \leq 2013} g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)$$ where $g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)=1$ if all $a_{i}$ are fixed points of $\sigma$ and 0 otherwise. (The $a_{i}$ 's need not be distinct.) Switching the order of summation, we find that the desired sum is $$\sum_{1 \leq a_{1}, a_{2}, a_{3}, a_{4} \leq 2013} \sum_{\sigma \in S} g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)$$ Note that the inner sum is equal to the number of permutations on $\{1,2, \ldots, 2013\}$ that fix $a_{1}, a_{2}, a_{3}$, and $a_{4}$. This depends on the number of distinct values the $a_{i}$ s take. If they take on exactly $k$ distinct values, then the inner sum will evaluate to $(2013-k)$ !, because $\sigma$ can be any permutation of the remaining $2013-k$ elements. (For example, if $a_{1}=a_{2}$ but $a_{1}, a_{3}$, and $a_{4}$ are distinct, the inner sum is 2010 ! because $\sigma$ can be any permutation that fixes $a_{1}, a_{3}$, and $a_{4}$.) Now, suppose we are given which of the $a_{i}$ are equal (for example, we could be given $a_{1}=a_{2}$ but $a_{1}, a_{3}, a_{4}$ mutually distinct, as per the above example). Assuming there are $k$ distinct values among the $a_{i}$, there are $2013(2013-1) \cdots(2013-k+1)$ ways to choose the $a_{i}$. At this point, there are $(2013-k)$ ! ways to choose $\sigma$ on the remaining $(2013-k)$ values such that it fixes the $a_{i}$, for a total of 2013! choices for $\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)$ such that $g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)=1$ and the $a_{i}$ satisfy the correct equality relations. Thus the answer is 2013 ! times the number of ways to choose equivalence classes on the $a_{i}$, so the problem reduces to finding the number of ways to partition 4 elements into nonempty sets. This process can be accelerated by doing casework based on the number of sets: (a) One set must contain all four elements, only one possibility. (i.e. all the $a_{i}$ s are equal) (b) Either one set contains 3 elements and the other contains the fourth (4 possibilities) or one set contains 2 elements and the other contains the other two (3 possibilities). (i.e. there are two distinct values of $a_{i}$ ) (c) One set contains two elements, the other two each contain one. There are $\binom{4}{2}=6$ ways to choose the two elements in the set with two elements, and this uniquely determines the partition. (i.e. there are three distinct values of $a_{i}$ ) (d) All sets contain one element, in which case there is only one possibility. (i.e. all the $a_{i}$ are distinct) Thus the number of ways to construct such a partition is $1+4+3+6+1=15$, and our answer is $15 \cdot 2013$ !