Elucidating reaction mechanism and purpose Please help me understand synthesis sequence shown below:

Benzylbromide is used for protection, probably from action of $\ce{CuBr2}$, but it's reaction mechanism is unknown for me. The bromination of the methyl group is some kind of radical reaction? What would it do to the rest of the molecule weren't it for benzylbromide? Next, 18-crown-6 is clearly used as catalyst for halogen exchange — but how does it function? Does it enclose bromine making $\ce{Br-}$ a better leaving group or $\ce{F-}$ better nucleophile? Finally, how does $\ce{CF3COOH}$ manage to remove that protecting group? Maybe it protonate benzilic $\ce{O}$ leaving the stabilized carbocation behind? But this seems quite unlikely since this would probably attacked fluorine or carbonyl.
I have no idea what the addition of benzyl bromide accomplishes. There doesn't seem to be any need for a protecting group in any of the subsequent reactions. The rest of the reactions are relatively easy to explain.

This paper reports essentially 100% yield for bromination of 4-hydroxyacetophenone with $\ce{CuBr2}$, refluxed in chloroform-ethyl acetate for 15-20 mins. There is no mention of any need for a protecting group for the phenol. No mechanism is given but you are correct in your assumption that radicals will be involved.

The second reaction is a regular $S_N2$ reaction. The transition state of the reaction is highly stabilized by the adjacent carbonyl group and so the reactivity is dramatically increased compared to a regular primary halide. 18-crown-6 is used to solvate the $\ce{K+}$ cations, for which it has a very high affinity. This leaves the fluoride free to react, making it far more nucleophilic than it would otherwise be. The reaction will go to completion easily because bromide is a far more stable leaving group than fluoride.

The final reaction is a standard ether cleavage, catalysed by a strong acid. The oxygen is protonated by the acid and leaves to reform the phenol and the stabilized benzyl cation.