Electronic – Meaning of infinite self inductance

I think your confusion lies in your first assumption. An ideal transformer doesn't even have windings, because it can't exist. Thus, it doesn't make sense to consider inductance, or leakage, or less than perfect coupling. All of these issues don't exist. An ideal transformer simply multiplies impedances by some constant. Power in will equal power out exactly, but the voltage:current ratio will be altered according to the turns ratio of the transformer.

For example, it is impossible to measure any difference between a 50Ω resistor, and a 12.5Ω resistor seen through an ideal transformer with a 2:1 turns ratio. This holds true for any load, including complex impedances.

^{simulate this circuit – Schematic created using CircuitLab}

Since an ideal transformer can't be realized, considering how it might work is a logical dead-end. It doesn't have to work because it is a purely theoretical concept used to simplify calculations.

The language you used in your first assumption is a description of the limiting case that defines an ideal transformer. Consider a simple transformer equivalent circuit:

^{simulate this circuit}

Of course, we can make a more complicated equivalent circuit according to how accurately we wish to model the non-ideal effects of a real transformer, but this one will do to illustrate the point. Remember also that XFMR1 represents an ideal transformer.

As the real transformer's winding resistance approaches zero, then R2 approaches 0Ω. In the limiting case of an ideal transformer where there is no winding resistance, then we can replace R2 with a short.

Likewise, as the leakage inductance approaches zero, L2 approaches 0H, and can be replaced with a short in the limiting case.

As the primary inductance approaches infinity, we can replace L1 with an open in the limiting case.

And so it goes for all the non-ideal effects we might model in a transformer. The ideal transformer has an infinitely large core that never saturates. As such, the ideal transformer even works at DC. The ideal transformer's windings have no distributed capacitance. And so on. After you've hit these limits (or in practice, approached them sufficiently close for your application for their effects to become negligible), you are left with just the ideal transformer, XFMR1.

Apparently, there is only one "leakage inductance" at one side in the figure above.

No, that is incorrect.

Every winding in a transformer does not couple 100% to each other winding and that is a fact. If some piece of text suggests that the leakage inductance is only attributable to one winding then that piece of text is at best misleading and, at worst blatantly wrong.

However, from the perspective of someone wishing to know how well two windings may couple then a single entity of leakage (a composite of both leakages) can be used to express that.

As for your 2nd question, you CANNOT short the secondary at the point you wish. This IS impossible - you can't take the equivalent circuit of the transformer and hack at it like that. The leakage measured is the composite leakage and this can be broken down into two components by using the turns ratio squared but even that is only an approximation; the magnetization inductance will alter the accuracy of this method slightly but, for all practical purposes, this method yields fairly accurate results.