Does fluorine ever form a double or triple bond?
The compound $\ce{NF}$, which is isoelectronic to $\ce{O2}$ is known, has been isolated in matrices and characterised and subjected to calculations. Like for dioxygen, three different states of this molecule are known: one triplet and two singlet states. Without performing any sophisticated analysis of its orbitals, we can expect a bond order of 2, and thus a double bond.

Harbison performed calculations on this compound. He came to the conclusion that the most stable triplet state is best described using only a single $\ce{N-F}$ bond. The two singlet states which require full electron pairing, however, display a much shorter $\ce{N-F}$ bond distance and are thus better described by an $\ce{N=F}$ double bond. Adding formal charges would lead to:
$$\ce{\overset{-}{N}=\overset{+}{F}}$$

Reference:

G. S. Harbison, J. Am. Chem. Soc., 2002, 124, 366–367. DOI: 10.1021/ja0159261.