I actually found the Klein language confusing. I had to read it several times to understand this. I read it as though induction was beating resonance, therefore we should get the product of induction. I kept thinking, that should be the meta-product. I could not understand how an inductive effect was "beating" resonance.
We are usually taught that deactivators are meta directors without being given much further explanation as to why that happens. What's really happening is that a deactivator, such as a nitro group with a positive formal charge, delocalizes the + around the ring in such a way that only the ortho and para positions are affected by such deactivation (that can be seen drawing the resonant structure for nitrobenzene for instance). These resonant structures are stable enough to be relevant and they are not in conflict with induction, which deactivates all the positions in the ring (and not just ortho and para, also meta). Meta (along with ortho and para) is deactivated by induction, but ortho and para are further deactivated by resonance as well. Therefore, the least deactivated position is meta (given that it has remained unaffected by resonance deactivation), and that's where electrophilic substitutions will tend to happen, at the least deactivated position. That's what usually happens with most deactivators, and why they are meta directing. Now let's come back to our exception: halogens.
Halogen substituents withdraw electron density from the ring by induction and, therefore, all the positions (ortho, meta and para) are deactivated (i.e. given a partial positive charge). Then comes resonance which, in this case, is the opposite to that which happens with the nitro group in that a negative charge is delocalized around the ring (affecting ortho and para, but not meta). Thus, the whole ring is deactivated by induction, but meta and para are activated by resonance. Therefore, the reaction will go ortho and para, the least deactivated positions. Now, at this point, you might ask: Why don't we put halogen substituents in the same category as we would put alcohols given that their behaviour is exactly the same (i.e. they are ortho,para orientators by resonance)? Why don't we simply consider them activators? In order to answer this question let's review how alcohols behave first.
Let's take the case of phenol (benzene with an alcohol substituent), whose -OH group is considered to be an activator. It draws electron density from the ring by induction (just like halogens) and therefore it deactivates the three positions (ortho, meta and para) giving them a partial positve charge by induction. Then, by resonance, it also behaves exactly like halogens, delocalizing a negative charge around the ring, which activates the ortho and para positions (with metha unaffected). The difference here is that the amount of negative charge put on ortho and para by resonance is of a higher "intensity" than the amount of partial positive charge put there by induction, therefore the net effect is that the ortho and para positions are more negative now than they would be if the -OH group was not present, because the negative charge given to them by resonant is enough to dwarf the delta+ charge given to them by induction; so, when all is said and done, ortho and para have gained electron density thanks to the -OH group, so we say that the ring has been activated at ortho and para, so the -OH group is considered an activator (even if it deactivates the meta position).
But, why doesn't the same thing happen with halogens? Because in the case of halogens the resonant structures that put the negative charge at ortho and para are so unstable that their "intensity" is not enough to offset the positive charge put on them by induction, so, when all is said and done, they have more of a positive charge than they would have without the halogen substituent's influence (i.e. they have been deactivated), and they will be less prone to electrophilic substitution than they would be in the case of benzene but, if/when a reaction has to happen, it will go ortho and para, since meta's situation is even worse (meta's been deactivated by induction but hasn't gained back even the slightest pinch of negative charge by resonance) .
So, in the end, an -OH group gives ortho and para more electron density than they would have in a unsubstitued benzene ring and so it's considered an activator, while a halogen gives ortho and para less electron density than they would have in a unsubstitued benzene ring and so it's considered a deactivator.
At last: why is resonance in the case of -OH enough to dwarf the inductive effects created by induction but not in the case of halogens? Because such resonance structures involve placing a + formal charge on the halogen and the halogen hates so so so much to be positive charged that it makes the resonance structures be not relevant enough to offset the delta+ charge.