Let $K$ be an $n$-manifold with boundary and let $x,y,z \in H^*(K)$ be cohomology classes with $x\cup y=y\cup z=0$.

The Massey product $\langle x,y,z \rangle$ is defined as the set of cohomology classes $[a\cup \tilde z - (-1)^{\deg (x)}\tilde x \cup b]$, where $a,b$ are cochains with $\partial a=\tilde x\cup\tilde y$, $ \partial b=\tilde y \cup\tilde z$ and $\tilde x,\tilde y, \tilde z$ are representatives of the cohomology classes $x,y,z$.

In his paper "higher order linking numbers" Massey uses another approach to compute elements of Massey products using the Poincaré-duality of cohomology+cup-products and homology+intersection of manifolds:

Let the fundamental classes $[M],[N],[P] \in H_*(K,\partial K)$ be the dual classes of the cohomology classes $x,y,z$, such that $X\cap P$ and $M \cap Y$ intersect transversally, for $X,Y$ manifolds with $M⫛ N= \partial X$ and $N ⫛ P= \partial Y$.

Then the sum $X \cap P - (-1)^{n-\deg(x)}M\cap Y$ obviously represents the Poincaré-dual of a triple product.

How exactly does this obvious duality work?

In my attempt I didnt get very far...

Using Bredon's intersection theory I get $[M \cap N]= D(x \cup y)$, for $D:H^*(K,\partial K) \to H_{n-*}(K)$ the duality isomorphism. This implies, that the cap product of a representative of $[K]$ and $\tilde x \cup \tilde y$ will be send to a representative of $[M \cap N]$, which we call $r_{mn}$. Since $x\cup y = 0$ it follows that $[M\cap N]=0$ and therefore $r_{m,n}$ is the boundary of a non-cyclic chain $r$.

Since this chain is non-cyclic, it is not a representative of any homology class. Thus I dont see any way to continue, using the duality $[M \cap N]= D(x \cup y)$.