This source states that the three s-orbitals of hydrogen and fluorine interact to form three new molecular orbitals, while other sources say that the 2s orbital is non-bonding.
Which one is more correct? Also, if they indeed form three new molecular orbitals, how do they look like?
Bearing in mind the energy and symmetry conditions, in the case of $\ce{HF}$ one can construct MOs using the $1s$ AO on H and the $2s$ and $2p$ AOs on F.
In general, the contribution to MOs is determined by the coefficients in the linear combination.
Here, one observes that the $1s$ electrons are almost completely localized on the $\ce{F}$ atom. Also, $1 \pi$ electrons are completely localised on the $\ce{F}$ atom because the $2p_x$ and $2p_y$ orbitals on F have a zero net overlap with the $1s$ orbital on $\ce{H}$.
Electrons in MOs localized on a single atom are referred to as nonbonding electrons.
Also, I would note that the $3 \sigma$ MO has less bonding character and the $4 \sigma^*$ MO has less anti-bonding character.
Note that the total bond order is approximately one because the $3 \sigma$ MO is largely localized on the F atom, the $3 \sigma$ MO is not totally bonding, and the $1\pi$ MOs are completely localized on the F atom.
On account of fluorine being a more electronegative atom, in the $2 \sigma$ bonding orbital the electron density is much greater on the more electronegative fluorine than on the hydrogen. However, in the anti-bonding $4 \sigma^*$ orbital, this polarity is reversed.
Caveat: The preceding paragraph may support your intuition and may be right in a few simple cases, but I wouldn't rely on it too heavily.
Given below is a diagram showing $2 \sigma$ , $3 \sigma$ and $1 \pi$ MOs in HF
