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I think the reaction is essentially an incomplete combustion reaction.

In a vaporizer, the idea is to get the propylene glycol, glycerine, and and "flavor" molecules into the vapor phase (or into aerosol droplets) without chemically degrading them.

However, the way this is accomplished is with very high-temperature resistor coils - the idea is that if you have a small amount of heat in a confined space, and the liquid is distributed throughout an absorbent (like cotton, for example), then the heat transfer will happen very quickly and a large amount of liquid will be vaporized.

The problem is that when you have a very high temperature heat source, and there is lots of oxygen in comparison to the liquid, there is a good chance that at least some of the liquid will reach the combustion temperature instead of just vaporizing.

What happens is that the hydrocarbons in the propylene glycol and glycerin molecules are partially oxidized - oxygen reacts with them to "steal" electrons. If the combustion was complete, you would wind up with just carbon dioxide and water. When combustion is incomplete, you can get any number of compounds, two of which happen to be formaldehyde and acetaldehyde. Carbon monoxide is another one I would be concerned about.

To see how this works, take a look at proplyene glycol (all pictures are from wikipedia):

and glycerine:

Notice how they both have three carbon atoms in the "backbone" and either one or none $\ce{O-H}$ groups (called hydroxyls or alcohols) attached to each one.

Oxygen can react with any one of those carbons and will try to form $\ce{CO2}$ and water, the most thermodynamically stable products that are possible:

If there is plenty of oxygen, it will. However, if there is not enough, you will get some intermediate species - formaldehyde or acetaldehyde, for example.

Let's look at formaldehyde:

In this case, oxygen has reacted with the the other two carbons on the backbone, leaving one with a hydroxyl. The hydrogen on the oxygen goes somewhere (either water or one of the other pieces of the original molecule) and the leftover electrons end up in the oxygen-carbon bond, forming a double bond. This molecule is more thermodynamically stable than proplyene glycol or glycerol, but less stable than carbon dioxide.

Now let's look at acetaldehyde:

Notice the similarity - it's almost the same, there is just a $\ce{CH3}$ (methyl) group in place of one of the hydrogens. You could get this by reacting the first carbon on propylene glycol with oxygen, and then removing the hydrogen from the hydroxyl on the second carbon. This is again more stable than propylene glycol, but less stable than carbon dioxide - it is also an intermediate.

Why does propylene glycol form more carbonylated products than glycerol?

This is a guess - but I think it likely has to do with the reaction kinetics. The reason I think this is true is that all of the intermediate products are less thermodynamically stable than carbon dioxide and water. Therefore, if a propylene glycol (PG) gives more of a given intermediate product than glycerine does, that product is "easier" to make for PG. This would happen if the activation energy for forming it is lower, or if the activation energy to go from the intermediate to carbon dioxide and water is higher compared to the intermediates formed by glycerine.

Note: Many people in other answers are pointing out that there are other mechanisms that could lead to these products, and that combustion is not likely at these temperatures. I think that given the mechanics of heat transfer we can't just rule it out, but I don't have any evidence for that. My answer is meant to show you how the reactant molecules contain the same structures as the products - I wouldn't put much faith in my mechanism (or anybody else's) without some real evidence.