The key idea is that $t \in T_{f}$ if and only if $f^{k}(t)=t$ for some $k>0$. To see this, let $s \in S$ and consider $$s, f(s), f(f(s)), \ldots, f^{2021}(s)$$ This sequence has 2022 terms that are all in $S$, so we must have a repeat. Suppose $f^{m}(s)=f^{n}(s)$ with $0 \leq n<m \leq 2021$. Then $f^{2021}(s)=f^{2021+m-n}(s)$. In particular, for $t=f^{2021}(s)$, we have $f^{k}(t)=t$ with $k=m-n$. On the other hand, if $f^{k}(t)=t$, then letting $s=f^{2021 k-2021}(t)$ gives $f^{2021}(s)=t$. We will compute the number of $f$ for which $f^{k}(1)=1$ for some $k$, and then multiply by 2021. We do this by casework on the minimum possible value of $k$. Given $k$, we just need to choose distinct values in $\{2, \ldots, 2021\}$ for each of $f^{1}(1), f^{2}(1), \ldots, f^{k-1}(1)$. We have $\frac{2020!}{(2021-k)!}$ ways to do this. For each of the $2021-k$ other values with $f$ not yet determined, we can do anything we want, giving $2021^{2021-k}$ choices. So, $$\sum_{f \in \mathcal{F}}\left|T_{f}\right|=2021 \sum_{k=1}^{2021} \frac{2020!}{(2021-k)!} \cdot 2021^{2021-k}$$ Taking this mod 2017, all terms with $k>4$ reduce to 0, and $2021^{2021-k}$ reduces to $4^{5-k}$ for $k \leq 4$. We are thus left with $$\sum_{f \in \mathcal{F}}\left|T_{f}\right| \equiv 4\left[4^{4}+3 \cdot 4^{3}+3 \cdot 2 \cdot 4^{2}+3 \cdot 2 \cdot 1 \cdot 4^{1}\right] \equiv 255 \quad(\bmod 2017)$$