Virtual $W^-$ boson may decay into $\bar{u}$ and $d$ quarks, $\bar{c}$ and $s$ quarks, $e$ and $\bar{\nu}_e$, $\mu$ and $\bar{\nu}_{\mu}$, $\tau$ and $\bar{\nu}_{\tau}$. Decay into $\bar{u}$ and $s$, $\bar{u}$ and $b$, $\bar{c}$ and $d$, $\bar{c}$ and $b$ are CKM suppressed. Also, as these hadrons ($\Lambda$, $\Xi$, $\Omega$ baryon, $K^{-}$) consist of $u$, $d$ &/or $s$ quarks, so decay of virtual $W^{-}$ boson (emitted by any of these quarks in these hadrons) into $\bar{c}$ and $s$ is kinematically forbidden but what makes the decay of this virtual $W$ boson into $\bar{u}$ and $d$ more probable than decay into $e^-$ and $\bar{\nu}_e$.
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I think that you are looking for the helicity suppression, this is simpler to understand for the decay of a $\pi$.
In leptonic decays you can separate the hadronic part and the leptonic one. You will get for the matrix element something like that: $ M \propto f_{\pi}p_{\mu}^{\pi}\bar{u}_{\nu_l}\gamma^{\mu}(1-\gamma_5)v_l \\ f_{\pi}\bar{u}_{\nu_l}\require{cancel}({\cancel{p}}_{\nu_l}+{\cancel{p}}_{l})(1-\gamma_5)v_l \\ \bar{u}_{\nu_l}\require{cancel}{\cancel{p}}_{\nu_l} = 0 ~ (\text{massless}) \\ {\cancel{p}}_{l}v_l = m_l v_l ~ (\text{lepton mass} ) \\ m_lf_{\pi}\bar{u}_{\nu_l}(1+\gamma_5)v_l $
The mass term produces the helicity suppression.
You can see that also thinking about the back to back decay, you have a left current, so the spins of the product would be directed in the same directions, but it's impossible because you have a pseudoscalar particle to start, so you have to do a spin flip but it costs the lepton mass term.