It is clear that there are $32$ domino tiles, hence there are exactly $32$ unit intervals that are covered by domino tiles thus they are internal for tiles and therefore they cannot be common for two tiles. The rest of unit intervals that are not at the edge of the chessboard are common for two tiles. Let their number be equal to $7 \cdot 8 \cdot 2 - 32 = 80$.
It is obvious that $2F + N = 80$, since the common segment for them has a length of $2$. Hence,
$$S = 2F + 2N = (2F + N) + N = 80 + N = 160 - 2F.$$

A set of tiles $D_1, \dots, D_k$, where $D_i$ has a tile $D_{i+1}$ are friendly for $i=1, k$, or a separate tile that has no friendly tiles, we call a chain. Each tile belongs to exactly one chain. Thus we have $C$ chains. Let it be $k_1$ tiles in the $l$-th chain. Then it has $(k_l-1)$ friendly pairs in total. Then there are $(k_1-1)+\dots+(k_C-1)=32-C=F$ friendly pairs in total. Therefore $S=160-2F=96+2C$. Thus in order to minimize $S$ we have to minimize the number of chains $C$. It is clear that there are no more than $8$ tiles in the chain, hence $C \ge 4$. Hence we have that the least possible sum is $S=96+2C=104$.
It remains to give an example when $C=4$. It is enough to place all the tiles in the same manner, for instance, the bigger side in all to be vertical.