Electronic – The Art of Electronics, figure 2.10B

I'm trying to learn electronics and right now I'm reading the book The Art of Electronics.

I'm stuck on this chapter 2 on figure 2.10B.

1. In particular I'm simply confused with the lingo like "pulling" collector to the ground, R3 "hold off" Q3, what does the author really mean? If it is some sort of quantity like voltage or current, why doesn't the author just say something like "voltage/current increases/decreases to X"?

2. What is "sinks the current"?

3. The part where it says "make sure you understand why?" well, no kidding I don't understand… 🙁 I've understood about NPN and PNP thing, like for example to get into saturation mode Vb > Ve and Vb > Vc (NPN) and stuff like that, but I still didn't get how R3 keeps Q3 off when Q2 is off?
   Well, I used the SPICE simulator, the result is exactly as the book says, but
   I don't understand why. Many book reviews seems to state the book is great for beginners, but it just skims through so fast on the foundations part.

Here is the text:

Figure 2.10B shows how to do that: NPN switch Q2 accepts the “logic-level” input of 0 V or +3 V, pulling its collector load to ground accordingly.
When Q2 is cut off, R3 holds Q3 off; when Q2 is saturated
(by a +3 V input), R2 sinks base current from Q3 to bring it
into saturation.
The “divider” formed by R2 R3 may be confusing: R3's
job is to keep Q3 off when Q2 is off; and when Q2 pulls
its collector low, most of its collector current comes from
Q3's base (because only ∼0.6 mA of the 4.4 mA collector
current comes from R3 – make sure you understand why).
That is, R3 does not have much effect on Q3's saturation.
Another way to say it is that the divider would sit at about
+11.6 V (rather than +14.4 V), were it not for Q3's base–
emitter diode, which consequently gets most of Q2's collector current. In any case, the value of R3 is not critical and could be made larger; the tradeoff is slower turn-off of Q3, owing to capacitive effects.