The diagram is correct. The dot markings indicate the polarity of the primary and secondary windings such that when instantaneous current enters the primary it comes out of the secondary as if the primary side circuit were directly connected to the secondary circuit.
Your primary has a voltage source vg, the secondary winding in this diagram will have vg present across it with the plus side at the top of the winding and the negative of vg being at the dotted end of the secondary transformer.
It depends what is connected to the other winding, which is why "mutual inductance" is also called "coupling factor" (They are not identical, but closely related terms).
The classic way of characterising a transformer's performance (after establishing n, the turns ratio) is to first measure the inductance of the primary - with the secondary open circuit. This measurement is the "primary inductance" - effectively unaffected by the other winding since no current flows in it.
And the primary inductance is an impedance connected across the power source - effectively wasted power, and as it is a low impedance at low frequency it determines the low frequency performance of the transformer.
Then re-measure the primary, but with the secondary short circuited. This is the "leakage inductance" (technically it's the parallel combination of primary and leakage inductances, but the primary inductance is usually a large enough impedance that it can be regarded as infinite, and ignored). Anyway the "leakage inductance" is essentially the coupling factor of the transformer into a short circuit - so in a good transformer it will be a very low impedance.
(The same pair of measurements can be made on the secondary, with the primary open/short circuit. It should give you the same result, scaled by n^2).
So the leakage inductance doesn't change the winding inductances - it couples one winding to the other, allowing the load impedance (scaled by 1/n^2) to appear in parallel with the winding inductance.
And the series combination of source impedance and primary inductance determine the LF response, while the series combination of leakage inductance and (load impedance/n^2) determine the HF response.
Best Answer
The first part of my answer was given when the OP specified micro henries rather than nano henries - that makes a big difference and so I would urge anyone reading this to bypass the first paragraph and concentrate on calculating the induced emf theory further below.
I doubt very much that 0603 inductors of values 20uH upwards will act like inductors much above tens of MHz due to their self resonant frequency turning them capacitive. Also inductors of this physical size and value will be extremely lossy.
However, if you still think you have a problem you could choose shielded types.
If you do want to calculate it then this might be useful: -
It tells you what the flux density is at a point on the axis of a coil distance "z" from the plane of that coil. You could plug in typical dimensions for an 0603 cross section and assume this is a circle of radius R.
It wouldn't be difficult to make a spreadsheet of results with z as the variable. Pretty soon after z exceeds R the flux density falls with z cubed so expect rapidly diminishing results.
To convert this to induced voltage you could reasonably assume that the flux density across the target 0603 component's cross section is constant and, using the cross sectional area, convert to flux.
You then have the fairly trivial formula V = N \$\dfrac{d\Phi}{dt}\$ to give you the induced voltage in the target component. However, the "usage" of the target component can make any induced voltage exaggerated or diminished and this is basically down to what other components attach to the 0603 target i.e. what type of circuit they are involved with.