Here is another short construction which is much simpler and just takes a few lines.

1. Let $M$ be a closed $n$-manifold. Consider the diagonal $M \to M \times M$. It is an embedding. Take its Pontryagin-Thom construction to get a map
$$
M_+ \wedge M_+ \to M^\tau
$$
(we have identified a tubular neighborhood of the diagonal with the total space of the tangent bundle).

2. If $M$ is (stably) framed, then $M^\tau \simeq M_+ \wedge S^n$ (stably). Then we have the map $M_+ \wedge S^n \to S^n$ induced by smashing $M_+ \to \text{pt}_+$ with $S^n$.  

3. The composition
$$
M_+ \wedge M_+ \to M^\tau \simeq M_+ \wedge S^n \to S^n
$$
is what you want: it's a duality map.

This can be seen on the level of homology (but it is enough to check a map is a duality map on the level of homology).