Selective transfer of least/most stable anion from organometallic reagents
It’s not really a dichotomy, it’s just two very different reaction mechanisms.

For the cuprate addition, which also applies to general Grignard-type additions, in principle the reaction is reversible (figure 1).


Figure 1: Reversibility of Michael additions.

However, the reverse direction is only really possible if the anion released is somewhat stable. In your example, thiophenolate may attack but may also be removed again in the course of the reaction. Once an alkyl chain has attacked, there will be no reverse reaction since the eliminiation of $\ce{R-}$ is too unfavoured.

For palladium-catalysed cross-coupling reactions, we need to consider the typical palladium catalysis cycle (figure 2).


Figure 2: typical catalytic cycle of palladium-catalysed ($\ce{M} = \ce{Pd}$) cross couplings. In a Stille coupling as in the question, $\ce{Y} = \ce{SnR3}$. Image taken from Wikipedia where a full list of authors is available.

The important step here is the transmetallation step: One of the ligands of metal $\ce{Y}$ ($\ce{Sn}$) must be transferred to the palladium ($\ce{M}$) centre. This will be the ligand that dissociates most easily — the best stabilised anion. So if the tin centre has three n-butyl ligands and one phenyl one, phenyl will be preferentially transferred since it dissociates more easily.

This is also part of the reason why $\ce{C}_\mathrm{sp^3}\ce{-C}_\mathrm{sp^3}$ cross coupling are so rare.