Just how do the underbanking and overbanking tendencies work?
I think your reasoning is correct. However, I would say that the change in lift due to effective AOA on each wing from yaw rate is an order of magnitude smaller than that from airspeed. At high bank angle, the wing loading is higher, resulting in more inward turn and decreased spiral stability. While the yaw-out moment from the drag is also increased due to relative airspeed difference on each wing, the effect is smaller since drag is smaller than lift.

Let's look at it more mathematically. Static spiral stability is dictated by:

$$C_{l_\beta}C_{n_r}-C_{l_r}C_{n_\beta}&gt;0$$

where $C_{l_\beta}$ is roll moment due to sideslip, must be negative for a Part 23 aircraft; $C_{n_\beta}$ is yaw moment due to sideslip, must be positive for Part 23; $C_{n_r}$ is yaw-out moment from yaw rate, where, as you reasoned, should be negative due to increased drag on the up wing and reduced drag on the down wing (due to airspeed difference from yaw rate); $C_{l_r}$ is the roll-in moment from yaw rate, where, as you reasoned, should be positive due to increased lift on the up wing and decreased lift on the down wing (due to airspeed difference).

At high bank angle, there is increased load factor (at 45deg, it's 1.4G). Therefore, there is increased lift and induced drag on each wing. The moment arm to induce the moment is about the same for each. However, if you look at the first order effect from the speed increase, it's:

$$\Delta F=\frac{1}{2}\rho (V+\Delta V)^2SC_x-\frac{1}{2}\rho V^2SC_x\approx\rho VSC_x\Delta V$$

So all things being equal, the larger the coefficient, the more pronounced the effect. And since lift is larger than drag, it wins out. (I'm still hand-waving here a bit since induced drag is squared of lift, so it's more nuanced).

With this in mind, you can see that at higher bank, $C_{l_r}$ increases more readily than $C_{n_r}$, therefore leading to reduced spiral stability.