Transformer magnetizing inductance
This is a great question Colin, one which I puzzled over long and hard myself before coming to an intuitive understanding.

Magnetizing inductance and magnetizing current are two aspects of the same departure of a real transformer from the ideal transformer model; in particular, as you mention, the fact that a real transformer core has finite permeability.

The permeability of an ideal transformer core approaches infinity. Interestingly, this means the magnetizing inductance of an ideal transformer is infinite, not zero. It does not appear in the ideal circuit because it models as an open circuit, not because it's not present. As a consequence, the flux needed in the ideal core to induce a voltage in the secondary is produced by a vanishingly small current (read zero). So the non-ideality of a real transformer is that it's inductance is less than infinite and therefore requires a current to flow to produce flux.

There are three other important factors to note about this:


The change in flux produces a voltage. It does not by itself produce a current. Essentially, the \$\frac{d\phi}{dt}\$ produces a voltage and the voltage produces a current if it can, based on the impedance of the secondary circuit.
For a given permeability and frequency, \$\frac{d\phi}{dt}\$ is proportional to the peak flux. If the peak (or peak-to-peak) flux is lowered, the voltage on the secondary drops. This means that the magnetizing current is the minimum current that can flow without dropping the output voltage. (Here I'm thinking of magnetizing current as sinusoidal AC. The same principles apply in switching transformers fed with square-wave input voltages but the complicating factors that introduces are perhaps better left until later.)
Nothing says the magnetizing current has to be in-phase with the input voltage, in fact, in a well-designed transformer, in general, the magnetizing current will be near 90-degrees out of phase. The real portion of the complex magnetic current gives a measure of the resistive and core losses. The imaginary part is pulsing back and forth, producing flux but not consuming power.


Now, when a load is placed on the secondary, a secondary current flows. This current produces a magnetic field (flux) that opposes that produced by the primary. This reduces the effective inductance of the primary which allows more current to flow through it, which produces more flux. This feedback cycle finds equilibrium when the magnetizing flux is restored and the output voltage is maintained. So the magnetizing flux is there at the same level at all times during the transformer's normal (i.e. non overload) operation. There might be a boat-load of other flux happening in both directions as power is transferred in a loaded state, but the net flux is always at the constant level produced by the magnetizing current.

From this you can see that the primary must always produce more flux than is cancelled out by current flow in the secondary and this surplus is provided by the magnetizing current. An open-circuit secondary takes the secondary current out of the picture and allows ready measurement of this magnetizing current.

This might make some sense of the expression "setting up the flux", although I think that's one of those less-than-instructive expressions that only makes sense when you already understand and don't need it any more :)