Is Fe(OH)2 soluble in alkalis?
This is a pretty interesting question. First of all, you are right, if you ever happened to see a yellow $\ce{Fe^3+}$ solution that was no pure aqua-complex but more likely $\ce{[Fe(H2O)5(OH)]^2-}$. The real $\ce{[Fe(H2O)6]^3+}$ for example, like it is formed in conc. $\ce{HNO3}$ or $\ce{HClO4}$ is actually somewhat purple. We call these compounds cationic acids. So the aqua-complexes will act as acids and become deprotonated in water. For your $\ce{Fe^2+}$ example it's pretty much the same. I often read in student reports that, if for example a hexaqua complex is treated with a base the $\ce{OH^-}$ would exchange for one water ligand. Many then concluded that $\ce{OH^-}$ had to be the stronger ligand in the spectrochemical series. This is, of course, wrong and not related to each other here. What is far more logical, at least to me is, that by coordination the water gets closer towards the highly charged metal center. But that means there is a repulsion from the $\ce{H}$ in $\ce{H2O}$ and the metal center since both have a positive (formal) charge. This makes it much easier to become deprotonated by a base. 

But why does this stop at some points? I asked myself the same question for the example of $\ce{BeO}$ reacting with $\ce{OH^-}$ to form $\ce{[Be(OH)4]^2-}$. Why is Beryllium oxide amphoteric forming a berrylate while $\ce{MgO}$ to $\ce{BaO}$ do not show this ability? It took me quite a while to find an answer to this. There is an equation by  Cartledge that tries to describe this phenomenon:

$$\ce{Φ = z/r}$$

Where Φ is the ionic potential, z the charge and r the radius. If $\ce{Φ^{1/2}}$ is smaller than 2.2 the ion is considered basic, above 3.2 it's considered to be acidic. In the middle, it's considered amphoteric. What does that mean? It seems like there is a ratio of radius to charge. Small, highly charged ions seem to polarize the surrounding water molecules so much that they quickly lose a proton. 

In your case, we are not specifically talking about oxides but I guess that something similar should be happening. The high charge on the metal and the fact that $\ce{Fe}$ is already in the middle of the d-row where atomic radii decrease along the row should be enough to actually cause even more deprotonation, this time perhaps not of surrounding water molecules like in the case of $\ce{BeO}$ but the ones that are already ligated to the metal center. At some point it will require additional help which happens when the pH is increased. And in fact for $\ce{Fe^2+}$ a  $\ce{Na4[Fe(OH)6]}$ seems to exist. 

I guess the comparison with the $\ce{BeO}$ may perhaps be a bit far fetched but it would be an actual example where you can see how the charge to radius ratio acts on the way soluble hydroxo-complexes are formed.