How is bismuth(III) able to form 3 bonds with hydroxide molecules in bismuth(III) hydroxide?
There isn't a unique sturcture for bismuth hydroxide, but ions such as

$$\ce{[Bi6O4(OH)4]^{6+}}$$

are involved, where the 6 Bi atoms are at the corners of an octahedron, none of the Bi atoms are directly bonded to each other, and O atoms bridge pairs of Bi atoms.

See Solvation of the Bismuth(III) Ion by Water... Inorg. Chem. 2000, 39, 4012-4021 as a starting point.  

But supposing the structure was $\ce{Bi(OH)3}$, Bi3+ has two electrons, each OH- has 8 electrons, so Bi could share 2 electrons from each OH- to form an octet.