The pion is the lightest hadron, a bound state of $\bar u$ and d, and a Goldstone boson of an axial flavor current with its isospin and parity quantum numbers. It is a pseudoscalar, so it is unlike the W - which is a vector. The $\rho^-$, also a hadron, couples to $\bar u ~d $ and is a vector, so it has the spin of W.
The W is over 500 times heavier than the π and couples to all quarks and all leptons. (Except in an extremely technical/fanciful Higgs-related way, it has no $\bar u ~d $ content.) The scale at which the W couples to the quarks is close to the mass of it (too technical), around 80GeV, and that mass enters by square in the denominator of the effective weak coupling constant (Fermi's coupling $G_F$) which is then substantially smaller than the strong-scale coupling the pion couples to these quarks with, called $f_\pi$, about 90 MeV.
At very low energies, the Goldstone coupling of the pion is crucially implicated in the chiral-symmetry breaking strong-interaction phenomenon that "converts" the quarks with mass 5-10MeV to "constituent quarks" of mass about 300MeV.
When the pion decays, it first converts to the pair of quarks you mention, and the pair of quarks further couple to the W which couples to μν, the decay products. It has to go to these, since there is no lighter hadron for it to decay to, strongly. So the so called "weak decay" (intermediated by a virtual W) is the only thing that can happen. And it happens proportionately to both $f_\pi$ and $G_F$ in the amplitude, so squared in the rate. This is where the name of $f_\pi$ comes from: it is measured in this decay and is characteristic of the pion, but part of the mysterious workings of nonperturbative QCD, and not really computable.
Lots of very different particles may couple to the same particles, with very different couplings indeed, and you hit on the one that virtually takes the prize in this sort of thing.