Are you sure you don't mean 'commute'? Monads, being endofunctors, compose just fine. Though the composition may or may not be a monad I suppose.
Types which are monads compose to make types that are definitely functors but maybe (probably?) not monads. That "maybe not" is what type theory people mean when they say "monads don't compose" but it's mostly not what's bugging people because...
Most monads can be rewritten as monad transformers (where applying them to the identity monad reproduces the original monad), which do compose in that way, and this is often what people do in Haskell when they have code that needs multiple effects.
But! Values of different types (like, say, a `WriterT t (Identity u)` and a `WriterT t (IO u)` can't be composed using the ever-so-convenient Monad interface which is the reason we wanted to make things Monads in the first place. This is what the article is discussing and it's what programmers often mean when they say "monads don't compose".
The workaround in the Haskell ecosystem (at least in those parts that haven't adopted effect systems or free monads w/ interpreters) is to write as much of your logic as possible using an unspecified monad, only asking for the functionality you need by way of constraints. Constraints compose like gangbusters! Values with unifiable types that have different sets of constraints? Well, it must've been a type that satisfies all of those constraints. And the abstract type is a nice place to stick in mocks as you test the pieces of things.
This shares much of the downside of effect sets, in that it is awkward to specify order ("layering"?) of effects when they do matter; it's possible to be more concrete about the types for those pieces but it does spread to the pieces that contain them and can sometimes get boilerplatey. Whether that's worth it for the added safety is going to be context dependent.
Maybe you already knew about this and is still asking, though?
If two monads commute you can show that the composition of the two is (trivially) a new monad, but I'm not sure if the converse also holds.
I started working it out by hand, but then figured nLab had it somewhere, and indeed: https://golem.ph.utexas.edu/category/2017/02/distributive_la...
F, G, and F.G are monads if and only if there is a "distributive law," which is a natural transformation G.F -> F.G satisfying some properties. It's like something that satisfies half the braiding properties, where braidings are already a weaker version of commutativity.
That said while FG = GF is indeed restrictive, requiring there to be a natural isomorphism between them is slightly less restrictive and just requiring the existence of a distributive law seems a bit too broad. What's preventing the existence of multiple distributive laws? Is there even anything preventing monads from always having a distributive law?
I don't know about whether distributive laws always exist, but what is probably true is that there's no universal distributive law -- that's in the sense that you have a function from pairs of monads to distributive laws that is natural with respect to homomorphisms of monads, whatever those are (I know there's a bicategory of monads, but that's about it).