How does power in "wall wart" power supplies work?
Yes, for a ideal power supply, power (current x voltage) in equals power out.  Of couse nothing is ideal, so the electrical power out will be less than the electrical power in with the difference wasted as heat.

Perhaps a example will illuminate this.  Let's say you have a 9 V 1 A wallwart.  That is what it puts out.  Under full load, the output power will be 9 W.  Let's say the wallwart is 80% efficient, so that means it will take 11.3 W from the outlet and the remaining 2.3 W will make it warm.  If the outlet is providing 110 V, then the wallwart will draw 11.3 W / 110 V = 102 mA from the outlet.  Hopefully this addresses the power in / power out part of your question.

As for the safety aspect, you seem to have some misconceptions.  First, the current rating of a constant-voltage power supply is only what it can put out, not what it will always force thru the load somehow.  If you stripped the wires from the 9 V 1 A wallwart and held the two ends in your hand, nothing much will happen.  That is because the resistance of your skin is high enough so that no dangerous currents will flow.  The fact the the wall wart could have produced 1 A doesn't matter.  Let's say the resistance of your hand between the two leads is 50 k&Omega;.  That means 9 V / 50 k&Omega; = 180 &micro;A will flow, which is not enough to harm your hand, probably not even enough to feel.