For two compounds that are symmetrical, why does the more compact one have a higher melting point?
The terms used by your teacher are quite in line with at least a qualitative explanation. 

I really suspect that there are various ways to express compactness and sort of "sphericity" of molecules. I am not a specialist so forgive me for being imprecise on this.

However, if we forget a rigorous classification in symmetry group, a potato is more spherical symmetrical than a banana.

In such a sense the ramified isomer is definitely more  symmetric and "round" than its linear counterpart. 

This is even more evident if all conformers are taken into account: in the linear molecule a chain end can even turn toward the center while the opposite being extended.

Now this does not necessarily explain the observed behaviour as for the more spherical molecule can pack easily but at expense of VdW forces that depend on molecular surface.

However, it relates to the fact that the linear isomer has a number of conformers much higher than the ramified one (rotation about 4 bonds versus 1 for each side of the ether)*** and results in the first having a bigger conformational entropic term.

Although the crystals will generally contain one isomer, the crystal formation take place at expense of the intramolecular forces. As such, when keeping entalpy terms comparable, bigger is the entropy term then less stronger is the binding.

$\Delta G = \Delta H - T\Delta S$,

For comparable enthalpies and very different conformational entropy terms, than the melting occurs at lower T for the entropy rich molecule, i.e. that one for wich more conformations become accessible. 

This obviously is not the case for the liquid to gas transition, as the conformational S term is similar in both phases and negligible faced to the gas entropic content. 

***Conformers due to the rotation around the bonds in the terbutyl groups do not really alter the shape of the molecules (connection with the compactness mentioned above). More rigorously at each side we should have (calling x the number of conformers around a bond) that the number of conformations is circa $x^4$ and about $3x^2$ for the linear and the branched molecules, respectively.

Additional note: conformational entropic terms are crucial to many phenomena in polymers and proteins (solubility, crystallization, protein folding/unfolding, etc).
A treatment of this requires statistics that I have forgotten. So take my numerical estimated about conformers with a grain of salt.  I am satisfied if I did convey the message. 

Add. Note 2. Another useful example is, for instance, a stiff and insoluble chain such as polythiophene.

It is insoluble as for electron delocalisation makes it flat and rigid. Once side substituted with alkyl chains, the entropic term brought in by the flexible chain is the main responsible for the acquired solubility.