Why is there no non-bonding molecular orbital in the overlap between H1s and Be2s in BeH2
Well, you need two sigma bonds here, and two anti-bonding ones. And there are actually two non-bonding orbitals here.

First of all: how you write it down doesn't really work, since there are p-orbitals in Be, not only s-orbitals. You need to look at them too, not only the s-orbitals. And I'm not quite sure if you did, but don't confuse Valence Bond and MO theory.

So for Valence Bond Theory:

We know this molecule is linear. Therefore we need two orbitals on the Be which can form bonds with the hydrogens and are in an 180° angle. So we do sp-hybridization and with one sp-orbital from Be and the s-orbital from hydrogen we can form a $\sigma$ and $\sigma^*$ orbital on each side. Two sigma bonds and two empty p-orbitals on Be which are non-bonding.

For MO Theory:

We construct a $\ce{H2}$ first, in this case we could write + + and + -, so we got a bonding and an antibonding orbital. Now we put our Be in the middle. And now it get's quite complicate to write down, so here on page 10 you can find the MO of $\ce{BeH2}$. You see that the bonding "$\ce{H2}$" orbital was combined with the 2s orbital of Be in an bonding and antibonding way, and the antibonding + - orbital of "$\ce{H2}$" was combined with a p-orbital in a binding and antibinding way. The two other p-orbitals aren't used and are non-bonding, since they would have equal constructive and deconstructive interaction with both "$\ce{H2}$" orbitals. In the end same result as with the valence bond theory.