Current in Transformer Primary (Forward Converter)
With the schematic posted as is, when Q1 turns off, in the ideal case, there is no current path available as you said. In the real world, a large positive voltage would appear at the drain of the Q1. Unless D1 breaks down first, the most likely current path is through Q1 under Vds breakdown. I think the circuit can actually work under this condition because a MOSFET can survive Vds breakdown if the power being dissipated is low enough.

By the way, infinite inductor means infinite impedance. Therefore an infinitesimal amount of current corresponds to a voltage approaching infinity. So the assumption such a (unloaded) transformer as being a short is not right.

In the real world, a common scheme of resetting the magnetizing current is by an extra reset winding on the transformer.

Going back to the diagram post, with the schematic as is, the current would not behave as illustrated. In order for the current to behave as illustrated, the transformer needs to be shorted by something when Q1 is off. One easy way to do something close to that is to put a diode (sometimes called a flyback diode in some applications) across the primary. Such a diode would not produce a working design in this application.