Let $X$ be a complex algebraic variety, $F$ a coherent sheaf. The question is whether $F$ admits a finite locally free resolution, and if not how to understand it with Poincaré duality.
If $X$ is not separated, it is possible that $F$ is even not a quotient of locally free sheaf, see MSE:850127. But if $X$ is smooth and separated then Hilbert syzygy theorem (or Hartshorne, exercise 3.6.9) implies that $F$ has a locally free resolution of length at most $\dim X$. Could you supply a counterexample for a separated but singular variety $X$, preferably normal and projective, such that there is clearly seen similarity with the topological failure of Poincaré duality?
Namely, let $K^0(X)$ and $K_0(X)$ be the Grothendieck groups of vector bundles and coherent sheaves respectively. Then there is a morphism $K^0(X) \to K_0(X)$ which is an algebraic analogue of the pairing with the fundamental class. It is surjective if and only if each coherent sheaf has a finite locally free resolution, and then it is also injective and is an analogue of Poincaré duality.