According to the Feynman diagram for $\phi$ meson decay, two gluons are required.
Why can a single gluon not be emitted from one of the initial-state quarks, creating the needed quarks to produce the two kaons?
This is seemingly possible for the $\rho$ decay. So why is it not possible for the $\phi$?
This answer doesn’t appear to address my question.
The question is not meaningful, as it succumbs to the myth/shared-metaphor of such light quark decays being describable by bona-fide Feynman diagrams, as opposed to schematic (Harari-Rosner ) diagrams tracking flavor.
Nobody has ever systematically computed such decays of light mesons in terms of perturbation theory in the large strong coupling, an impossible QCD task. (Alternatively, the blue pancakes in your shared-metaphor pictures summarize exchanges of an indefinite number of gluons, which conflate 1 with 2, with 3 .... "gluon" exchanges. Check there are no color bottlenecks involved!)
In the early 70s, people kept track of flavor and fermion number of valence quarks in hadrons sort of like this, skipping "gluons", as hadrons are basically impossible "wave functions" of valence quarks and an infinity of gluons and quark-antiquark "sea" pairs. With the advent of QCD, misguided well-meaning teachers, often experimentalists, started adding gluons to morph such diagrams into Feynman-like ones, with might appear more palatable to large uninformed audiences.
With the advent of heavy quarks, like c, b, and t, involving large momentum transfers and thus smaller strong couplings, such diagrams were less preposterous, and became closer to 0th order Feynman ones, so these pictures were established, however vague, in cartooning light-quark (u,d,s) hadrons, used here.
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