How do alkenes react with bromine?

Alkenes perturb the electron cloud in #"Br"_2#, polarizing it and allowing addition of #"Br"^(+)# across the alkene.

The remaining #"Br"^(-)# can backside-attack to form the vicinal dibromide product.

1-1. The #pi# bonding electrons from the alkene move towards one #"Br"#'s antibonding orbitals; that #"Br"# becomes partially positive (#delta^(+)#), and the top #"Br"# becomes partially negative (#delta^(-)#).

This accounts for the right-hand arrow.

1-2. Polarizing a bond by perturbing the electron cloud weakens the bond, and in this case, it was enough to break the #"Br"-"Br"# bond, and the bonding electrons move into now-nonbonding orbitals.

This accounts for the upper arrow.

1-3. However, the bottom #"Br"# also donates electron density into the alkene's antibonding orbitals, thus making a bridging connection between the left and right carbons.

This accounts for the left-hand arrow.

That was the complicated part. The rest is not too crazy.

2. The remaining #"Br"^(-)# is the nucleophile, attacking one of the carbons from the rear (a backside-attack), and breaking one of the bridging bonds (which are weak already).

This accounts for both arrows.

The backside-attack is what generates an anti-addition product. This is reflected in the trans relationship of both #"Br"# on the final product.

In terms of Markovnikov or anti-Markonivkov addition, it doesn't really matter because both #"Br"# added are identical atoms.