(I have heard people say that the cup product is commutative up to homotopy)

It takes more than just being commutative up to homotopy.

If x and y are cochains, then there must be some homotopy F from xuy to yux (where I'm using "u" for the cup product). However, if F' is some other homotopy from xuy to yux, then there must be a homotopy G from F to F'.

Then if G' is another homotopy from F to F', then there must be a homotopy H from G to G'.

Then if H' is another homotopy from G to G', then...

The reason this is so is probably best illustrated by noticing that simplices are contractible.

(For what's blow, the diagram here near the top of the "Interpretations" section shows the two maps.)

Consider the 1-dimensional level. If I and I' are 1-simplices, then the product simplex is IxI', which you can imagine as a rectangle. There is a corresponding diagoanal leading from the pair of left endpoints to the pair of right endpoints. This is what the cup product "should" be in the sense that the diagonal map becomes the cup product after taking cohomology.

However, what's described above isn't actually a product since it goes from C*(X x X) to C*(X), where we really want a map whose domain is C*(X) tensor C*(X). So we need an approximation map. Again, we'll go back to the level of simplices, so I want to construct the map as S(X x X) \to S(X) tensor S(X), where S(-) denotes the singular complex.

Back to the level of a product of one-simplices, the thing I have in my product complex is that diagonal of the square. However, this isn't a tensor product of simplices. For the tensor product, one-dimensonal things are products of one 0-dim and one 1-dim, i.e., the edges of the square.

So you're trying to get from one corner of the square to the other but you have to go around the outside. Well, there are two ways of doing this, and they are the two orders of the product xuy and yux. Each path is homotopic to the diagonal (telling you it's the "correct" thing) and so each is homotopic to the other (so the square gives a homotopy).

Now if you have two such homotopies, you have a product of squares, and again can get from one to the other in two ways, but the product of two squares is contractible so there's a homotopy between the two ways. (These homotopies of homotopies are the "cup-i" products, and give rise to Steenrod operations.)

A similar story happens in higher dimensions: if you have an n-simplex and an m-simplex, you want the diagonal into the (m+n)-dimensional product, but that diagonal isn't in the tensor product so you have to approximate by going around edges, but both choices are homotopic since the whole thing is contractible, and so on.