Why does a foster seeley circuit need transformer coupling?
You have already described that you need loose coupling in order to get a 90 degree phase shift between the primary and the secondary. The phase shift will vary with the deviation of the primary frequency from the resonance frequency of the secondary. It is only 90 degrees right at resonance.

But you are wrongly assuming that the energy that makes the secondary tank circuit oscillate, is fed to it via Cc and the center tap of the secondary. This is untrue. It is the loose coupling that provides the energy transfer from primary to secondary. The optimal amount of coupling depends on the Q of the secondary tank. The higher the Q, the looser the coupling.

The center tap, and the voltage coupled into it via Cc, acts as a phase reference, as @Buck8pe has written. If you look at just the top half of the secondary circuit, consisting of the top half of the secondary winding between center and top, and the corresponding rectifier circuit, then you realize that the anode of the rectifier sees the sum of the primary voltage (via Cc), and half the secondary voltage (via the secondary winding). This means you are adding two sinusoidal voltages with the same frequency, but a phase difference between them. The resulting signal is again a sinusoid with said frequency, but with an amplitude that depends on the phase difference. Hence you have converted a phase difference into an amplitude variation, which you can detect using the rectifier/capacitor combination, as usual.

The bottom half does the same, except that the inverted secondary voltage is used, so you get the difference rather than the sum of the two sinusoids. Combining the two has the effect of linearizing the demodulation, hence reducing distortion.