Let $M$ be a closed orientable $n$-manifold containing the compact set $X$.
Given an $n-q-1$-cocyle on $X$ (I am choosing this degree just to match with the notation
of the Wikipedia article to which you linked), we extend it to some small open neighbourhood $U$ of $X$.
By Lefschetz--Poincare duality on the open manifold $U$, we can convert this $n-q-1$-cocylce
into a Borel--Moore cycle (i.e. a locally-finite cycle made up of infinitely many simplices)
on $U$ of degree $q+1$. Throwing away those simplices lying in $U \setminus X$,
we obtain a usual (i.e. finitely supported) cycle giving a class in $H_{q+1}(U,U\setminus X) = H_{q+1}(M,M\setminus X)$ (the isomorphism holding via excision).
Alexander duality for an arbitrary manifold then states that
the map $H^{n-q-1}(X) \to H_{q+1}(M,M \setminus X)$ is an isomorphism. (If $X$ is very pathological, then we should be careful in how define the left-hand side, to be sure
that every cochain actually extends to some neighbourhood of $X$.)

Now if $M = S^{n+1}$, then $H^i(S^{n+1})$ is almost always zero, and so we may use the boundary map for the long exact sequence of a pair to
identify $H_{q+1}(S^{n+1}, S^{n+1}\setminus X)$ with $H_{q}(S^{n+1}\setminus X)$ modulo worrying about reduced vs. usual
homology/cohomology (to deal with the fact that $H^i(S^{n+1})$ is *non-zero* at the extremal
points $i = 0$ or $n$).

So, in short: we take a cocycle on $X$, expand it slightly to a cocyle on $U$,
represent this by a Borel--Moore cycle of the appropriate degree, throw away those simplices lying entirely outside $X$, so that it is now a chain with boundary lying outside $X$, and finally *take this boundary*, which is now a cycle
in $S^{n+1} \setminus X$.

(I found these notes of Jesper Moller helpful in understanding the general structure of Alexander duality.)

One last thing: it might help to think this through in the case of a circle embedded in $S^2$. We should thicken the circle up slightly to an embedded strip. If we then take our cohomology class to be the generator of $H^1(S^1)$, the corresponding Borel--Moore cycle is just a longitudinal ray of the strip (i.e. if the strip is $S^1 \times I$, where $I$ is an open
interval, then the Borel--More cycle is just $\{\text{point}\} \times I$).

If we cut $I$ down to a closed subinterval $I'$ and then take its boundary, we get a pair
of points, which you can see intuitively will lie one in each of the components
of the complement of the $S^1$ in $S^2$.

More rigorously, Alexander duality will show that these two points generate the reduced $H^0$ of the complement of the $S^1$, and this is how Alexander duality proves the Jordan curve theorem. Hopefully the above sketch supplies some geometric intuition to this argument.