The asymptotic phase behavior of an RHP zero is from 0 degrees to -90°, the mirror of an LHP zero. Graphically, I'm confused about why this is the case and the phase is not from +180° to +90°. See the below image for my reasoning/confusion. Is the angle definition not consistent?
You're right, a single (real-valued) zero in the RHP could cause a phase that moves from $\pi$ at $\omega=0$ to $\pi/2$ for $\omega\to\infty$. However, if you rewrite the corresponding contribution to the transfer function as $s_0-s$ (i.e., you change the sign), then you obtain a phase shift between $0$ at $\omega=0$ and $-\pi/2$ for $\omega\to\infty$.
The difference is just a sign inversion, i.e., an addition of $\pm\pi$ to the phase. The contribution of a single real-valued RHP zero in the form $s_0-s$ does not invert the sign for $s=0$, whereas if you write it as $s-s_0$ you get a sign inversion, hence the additional phase shift of $\pi$.