Which lone pairs/bonding pairs go into which pi molecular orbitals?
This is a common attempt to visualise what &#x3c0; systems actually mean for the overall Lewis structure. But let me note a couple of things first:


Obviously, there is no $\mathrm{sp^3}$ lone pair on DMF’s nitrogen. Instead, the atom is $\mathrm{sp^2}$ hybridised and the lone pair (if it is there) occupies a p-type orbital.
Remember that the &#x3c0; analysis only deals with the &#x3c0; orbitals (obviously); there is an underlying &#x3c3; system which is implied but hardly mentioned. Thus, if ‘one bond’ is created, it is actually a double bond, while ‘no bond’ means ‘only the underlying single bond’.




If you want, in simplistic terms, you can add the &#x3c0; orbitals one after the other. The first is fully bonding and has contributions by all three atoms. Since it is fully bonding across two bonds but only contains two electrons, we can say that it contributes $\frac22$ bonding electrons to each bond or a bond order of ½.

The second orbital is not really bonding as it has a perpendicular nodal plane through the central atom (carbon). Thus, we cannot really analyse it in terms of the $\ce{C-O}$ or $\ce{C-N}$ bond order directly. However we can say that the electron density will be stronger at the ends, i.e. on nitrogen and oxygen. Each of these atoms practically experiences $\frac22$ lone pair electrons (do not participate in bonding) or ½ a lone pair.

Consider what would happen if you had filled the third orbital. It would have been antibonding with respect to both the $\ce{C-O}$ and the $\ce{C-N}$ bond. It would have contributed the same bond order of ½ to both — but this time in negative. That would contribute to an overall bond order of zero or the bonding electrons being on either side of the bond as lone pairs.

For your question, this means:


the fully bonding orbital cannot contribute to any lone pairs since it is not compensated for by an antibonding one
the nonbonding orbital contributes the lone pair. Since it only experiences contributions from oxygen and nitrogen, this is where the lone pair will be located.
it would be wrong to look at the Lewis structure, see a lone pair on nitrogen and immediately associate that with the ‘lone pair’ of the question. If anything, consider all relevant resonance forms and all their lone pairs. Also, as soon as it has resonanted to form a bond, it is no longer a lone pair.
finally, as per your side question, note that the $\ce{C-H}$ bond in DMF has a very, very low tendency to deprotonate. That proton is not even mentioned on both the Evans table and the Bordwell data. While it might lead to a carbene, the carbene is not exactly stable. A much more likely deprotonation would be that of a methyl group which would extend the &#x3c0; system. Note that the sequential treatment with butyllithium and DMF is a method to convert electron-rich aromatic rings (e.g. thiophene) into their corresponding carbaldehydes — in this reaction, the proton of DMF is less acidic than that of a thiophene or proton exchange would occur in place of nucleophilic attacks.