I don't know if I fully grasp your question; but I can confirm that slowness and velocity vectors are kind of weird things :-)
Let me describe a simple example (and not worry too much about the mathematical details). If in the figure above, if you point your wave not diagonally but just straight down, what happens? Well, $dx$ becomes infinitely large when the incidence angle becomes $i=0^\circ$, because there is no observable thing happening along the upper horizontal surface, the wave only moves down! At the same time, $dz$ becomes exactly equal to $ds$ when the incidence angle $i=0^\circ$. So, $dx=\infty$ while $dz= ds$, right? I have drawn that in the figure below, for a monochromatic (single-frequency) wave, such that for all the 3 cases, the length of the vector $\vec{p}$ is identical! I have added $c_x$ and $c_z$ as the apparent velocities.
So for a wave moving straight down, the wave moves in 2D for a grand total distance of $ds$, it propagated 0 in the direction of $dx$, and $ds$ in the direction of $dz$. Now your research already suggests that either of the following relations should hold -- both basically expressions of the Pythagorean rule: $$ dx^2 + dz^2 = ds^2, \tag{a} $$ or $$ \frac{1}{dx^2} + \frac{1}{dz^2} = \frac{1}{ds^2} \tag{b} $$ If we use our extreme example $dx=\infty$ and $dz=ds$, then only equation b can hold (remember, $1/\infty=0$).
If we multiply every term in equation $b$ with $dt^2$, we get: $$ \frac{dt^2}{dx^2} + \frac{dt^2}{dz^2} = \frac{dt^2}{ds^2}, \tag{b'} $$ or realizing that $dx/dt=c_x$ and $dt/dx=1/c_x=p_x$ we get: $$ p_x^2 + p_z^2 = p^2 = \frac{1}{c^2}, \tag{b"} $$
With those details out of the way, we have seen the following: $p_x=0$ but $c_x=\infty$, and $p_z=1/c_z$ and $c_z=c$. 453
So we have found that both ${c_x,c_z}\geq c$, i.e., both apparent velocities were greater than or equal to the original wave speed! On the other hand, ${p_x,p_z}\leq p$, i.e., the apparent slowness components were less than or equal to the original slowness.
And that is the result that holds generally. The apparent velocity components are always equal to or greater than the original velocity; while the apparent slowness components are always equal to or less than the original slowness. This is why equations (b), (b') and (b") give the right decomposition rule, it's the only one where $p_x$ and $p_y$ are less than the full $p$! This is why we can think of such a thing as a slowness vector $\vec{p}=(p_x,p_y)$, because the modulus of that vector $\lvert \vec{p} \rvert^2 = p_x^2+p_y^2$ (which is eq. (b")) behaves exactly like a vector should!
I hope that maybe along the way I answered your question somehow? My apologies for the rather mathematical take on things, but it's the only way I ever understood slowness vectors! 499