OK, let's step through your questions one at a time.
First off, what do they mean by steady-state number here?
Intermediates, when formed in reactions, usually exist in low concentration. The steady-state assumption, when applied to reaction kinetics, is assuming that the concentration of (this low-concentration) intermediate will not change. The rate of its production is equal to the rate of it's consumption. The steady state assumption allows us to recast the previous statement as a mathematical equation. Appling the steady state assumption to the problem at hand allows us to state that for all of the intermediates (dust particles with 1 or 2 hydrogen atoms on them) $$\ce{the~ rate~ of~ hydrogen~ atom~ adsorption = the~ rate~ of~ hydrogen~ atom~ desorption}$$
in this case why must the rate of absorption be equal to the rate of desorption
or, $$\ce{k_a [H]=k_{d} N}$$
why is it $\ce{k_{d} N}$ ?
Because $\ce{k_{d} N}$ represents the Total rate of desorption of hydrogen atoms from dust particles. it is equal to $\ce{k_{d} x_{1} + 2 k_{d} x_{2}}$
Further, why is the rate of desorption not simply the reverse of the absorption reaction and thus the same rate but with a negative sign?
It is the same rate, that's what $\ce{k_a [H]=k_{d} N}$ means. Rates are always positive, we can insert a minus sign later when we set up an equation.
how did they come up with those rates?
- Adsorption: rate= $\ce{k_{a}{H} x_{m}= K_{a} [H] x_{0} + K_{a} [H] x_{1}}$, it is just the typical kinetic expression - adsorption rate x concentration of hydrogen x concentration of the various dust particles that can adsorb a hydrogen atom
- Desorption: rate = $\ce{(k_{d} * 0 * m_{0}) + (k_{d} * 1 * m_{1}) + (k_{d} * 2 * m_{2})}$, "m" is a statistical factor that represents the number of hydrogen atoms on a dust particle, it can be 0, 1 or 2.
- Reaction: rate = $\ce{(1/2 * k_{r} * 2*1 * x_{2}) + (1/2 * k_{r} * 1*0 * x_{1})}=k_{r}*x_{2}$
Why does absorption have $\ce{[H]}x_m$ but desorption have $\ce{m}x_m$?
As noted above, "m" is a statistical factor. For example desorption can occur twice as frequently from $\ce{x_{2}}$ as from $\ce{x_{1}}$, "m" corrects for this. Adsorption needs no statistical correction, one hydrogen atom is adsorbed or it isn't.
Finally, for the reaction rate, where did they get m(m−1)?
See how I used it above. It is used in the reaction rate as a statistical factor and also to cause terms such as $\ce{(1/2 * k_{r} * 1*0 * x_{1})}$ to go to zero since $\ce{m_{1}}$ can't react.
why does the $k_a$ term have [H] while the other terms do not?
Hydrogen is used up (consumed, e.g. it is adsorbed, taken out of circulation) only in the adsorption step, therefor its concentration must show up in the rate expression for the adsorption step.
What happened to the $m x_m$ that was mentioned above?
Since the full rate expressions have been written out, they have been replaced with integers 0, 1 and 2 like I did above.
Further, how did they derive the rate of change for $x_1$ and $x_2$?
rate of change of $\ce{x_{1}}= rate of formation - rate of destruction
the rate of formation is given by the rate of $\ce{x_{0}}$ adsorbing a hydrogen atom = $\ce{k_{a} x_{0}}$ plus the rate of $\ce{x_{2}}$ losing a hydrogen atom = $\ce{2*k_{d}x_{2}}$
the rate of destruction is given by the rate of $\ce{x_{1}}$ adsorbing another hydrogen atom or losing a hydrogen atom = $\ce{k_{a}[H] x_{1} + k_{d}x_{1}}$
For the time being, I'll leave the rate of change of $x_2$ as an exercise for you to test yourself on what you've learned.
I need to head out now, I'll proof this tomorrow.