Fixed – thank you very much

There’s nothing to say that they *can’t*, specifically, although with alkenes, the precursors of bromine and chlorine radicals also tend to add as electrophiles to double bonds (e.g. forming bromonium and chloronium ions) which would not be good for polymerization. Peroxides don’t tend to have these cationic pathways.

The reaction will favor breaking of C-H over C-Cl bonds because the resulting H-Cl bond (103 kcal/mol) is stronger than the resulting O-Cl bond (65 kcal/mol) . The O-Cl bond is quite weak due to repulsion between the lone pairs. The driving force for a radical breaking C-Cl is not strong enough.

I don’t know how to answer your question directly. If you are asking, “what determines what product will be formed”, the answer lies in comparing the activation energies for the various competing rate-determining steps.

It’s worth recalling that atomic carbon, C , has four valence electrons, four empty places in its octet, and is neutral.
Atomic fluorine, F• , has seven electrons, one electron away from a full octet, and is also neutral.
Just because a species is reactive and has a high electron affinity does not mean that it has to be charged.
For proof, go to the formal charge formula for the methyl radical, H3C• . If carbon has 4 electrons to itself, it is neutral. Here it has one electron to itself (the radical) and has a 50% share in the six bonding electrons of the C-H bonds, (3) giving a total of 4. If it loses the electron it becomes the methyl cation, CH3(+). If it gains an electron it becomes the methyl anion, CH3(-).

1° is a primary free radial where the carbon bearing the radical is attached to one other carbon. 2° is a secondary free radical where the carbon bearing the radical is attached to two other carbons.