It looks like you are understanding correctly.
As the current increases the flux builds up in both inductors, which you correctly sum together giving a slope of 2 webers/amp. That is until inductor 1 saturates and no longer does increased current produce increased flux. At that point on the total flux will only increases from inductor 2 at 1 weber/amp. The behavior is similar when the current is reversed, as you have correctly shown.
I'm just guessing the units are weber/amp.
I don't blame you for stumbling on the paralell case. In esence your being asked to imagine what two ideal inductors would do if supplied with a DC constant current source. Try that in a simulator I bet you will get an error.
With that in mind lets think about parallel circuits in general.
Remember that the total impedance of a parallel circuit is less than any one element of the parallel circuit.
To help understand what effects that can have I like to think of the following.
Imagine you connect in parallel a 10 ohm resistor in parallel to a 10 megaohm resistor. The total impedance is just shy of 10 ohms. It's almost as if the 10megaohm resistor isn't there and the 10ohm resistor "dominates" the share of the total current.
So let's relate that back to the amount of flux problem you have. My guess is the total flux will only change at the same rate as the "slowest" inductor. So looking at increasing positive current, the flux will increase at 1 weber/amp until one amp of current. From there the flux will no longer increase as inductor 1 is now saturating. That saturation "dominates" the total inductance of the circuit.