Your description of the six functors does not mention any relations between the $!$-functors and the $*$-functors or the tensor product, which is where these dualities are hiding.
Poincaré duality is a relation between the $*$-functors and the $!$-functors. Typically there are canonical isomorphisms $f_!=f_*$ when $f$ is proper and $f^!=f^*[d]$ when $f$ is nice (e.g. smooth) of relative dimension $d$. Depending on how the $!$-functors are constructed, one of these isomorphisms will hold by definition but the other one will be nontrivial to prove. When $f:X\to S$ is nice, we get isomorphisms (where $1_S$ is the unit object in $D(S)$) $$ H^n(X,M) := \mathrm{hom}(1_S, f_*f^*(M)[n]) = \mathrm{hom}(1_S,f_*f^!(M)[n-d])=:H_{d-n}^\mathrm{BM}(X,M) $$ and $$ H^n_c(X,M):= \mathrm{hom}(1_S, f_!f^*(M)[n]) = \mathrm{hom}(1_S,f_!f^!(M)[n-d])=:H_{d-n}(X,M), $$ On the other hand, there are horizontal identifications when $f$ is proper.
Serre duality uses a relation between the $!$-functors and the tensor product. Namely, for $f:X\to S$, the functor $f_!$ is $D(S)$-linear; in particular we have the projection formula $f_!(A\otimes f^*(B))= f_!(A)\otimes B$. By adjunction this implies the isomorphism $$ f_*\mathrm{Hom}(A, f^!(B)) = \mathrm{Hom}(f_!(A),B). $$ One gets Serre duality when $f$ is proper (so $f_!=f_*$) and $B=1_S$ (so $f^!(B)$ is the "dualizing sheaf"). Combining the projection formula with Poincaré duality we can also deduce that $f_*$ preserves $\otimes$-dualizable objects when $f$ is both proper and nice, which gives a perfect pairing. But the usual formulation of Serre duality in terms of cohomology is specific to the derived category of quasi-coherent sheaves when $S=\mathrm{Spec}(k)$ is a field (note that in the quasi-coherent case the adjunction $(f_!,f^!)$ only exists when $f$ is proper).
There are further duality statements, such as Verdier duality, which says that $\mathrm{Hom}(-,f^!(1_S))$ is an equivalence of a certain subcategory $D_c(X)\subset D(X)$ of "constructible"/"coherent" objects with its opposite. But unlike the previous two, this statement does not directly follow from the various relations between the six functors.