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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.

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The simplest example is $X = Spec(k[t]/t^2)$ and $F = k$.

The simplest normal example is the quadratic cone $X = \{xz -y^2\} \subset \mathbb{P}^3$ and $F$ is the structure sheaf of the vertex of the cone.

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    The 1st example is explained with details as Example 4.18 in this notes 'An Introduction to K-theory' by Eric M. Friedlander here: http://users.ictp.it/~pub_off/lectures/lns023/Friedlander/Friedlander.pdf – 2017-02-09