Some questions about a magnetically coupled circuit
The first case (k=1, C1=C2=1uF) can be considered as:



(See the ADDENDUM for a quick proof)

After the initial impulse there's 2 lossy capacitors in parallel with the inductor. (NOTE:the voltage source is like a short circuit after the impulse has gone) That's a resonant circuit with 2uF effective capacitance. The resonant frequency is about 1100Hz. The oscillation dies away due the losses in the resistances. C1=C2, so the current distributes equally between them.

In your middle case (k=1, C1=1uF, C2=1000uF) the equivalent circuit is the same except C2=1000uF. It practically with 10mH inductance defines the resonant frequency which is now about 35Hz. During the oscillation after the initial impulse the current mostly goes trough C2 because there's virtually the same AC voltage over the capacitors, but the reactance of C2 is much lower. During the initial impulse C2 doesn't get much voltage because the capacitive  voltage division leaves most of the initial impulse voltage to C1 and the inductor does not charge much current during the impulse due the low voltage over it during the impulse. As you see the oscillation current is mostly through R2 i.e. C2 and it's much lower than the current in the first case. In the screenshot you can see about 1 pixel high amplitude.

In your third case (k=0,01 C1=1uF, C2=1000uF) we have two only loosely coupled resonant circuits with different resonant frequencies. After the initial impulse the leftmost resonator oscillates and part of the AC current is connected to the other resonator. It's far from the resonance, so no complex interactions occur between the resonators.  The right side gets some AC due the coupling but the current is much lower than in the left side.

You can think that in case 3 the right side coil generates an AC voltage, which is practically shorted by C2. The current at the right side is practically k * the current in the right side, as you can easily see if you can form the mutual inductance relations with complex phasors.

ADDENDUM: The reduction of the inductors to one can be derived by using the ordinary mutual induction relations: