How is bismuth(III) able to form 3 bonds with hydroxide molecules in bismuth(III) hydroxide? Bismuth(III), if I am not mistaken, has only two valence electrons in the 6s orbital. However it is able to form 3 bonds with hydroxide. This goes against my intuition. How is it able to form 3 bonds with 3 hydroxide molecules with only 2 valence electrons? 

my best guess is that i am mistaken in thinking that the bismuth(III) is forming 3 covalent bonds when in actually it is forming ionic bonds. As a result, bismuth does not need to offer/share 3 electrons to each hydroxide, but simply take 3. Is this correct?
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.