Preference of certain positions during chlorination of an alkene
In your first example, propene plus chlorine, one of the Cl atoms bonded to a "CH" and the other, that you described as bonding to the end, bonded to a "CH2," so it's not just CH groups getting all of the chlorines.  Instead, it's the carbons at the ends of the double bond, however many hydrogens they might be attached to.

I think part of your misunderstanding lies in the way that you're writing/imagining the alkene.  The propene molecule has a shape, and the geometry of the molecule plays a major role in its reactions.  The double bonded section is more or less flat, and the CH3 is spread out in space.  The chlorine atoms attach to the ends of the double bond, wherever that may be.  The mechanism for this attachment is beyond the scope of your question, but if you're curious, you can find more information here.  At the very least, know that the chlorines do not attach to the carbons at the same time; there is an intermediate compound formed.



(If my drawing isn't clear, you can try building a model of the molecule.)

It would require the breaking of a C-H or C-C sigma bond in order for the chlorine to get to the CH3 carbon, and that's unlikely.  As I mentioned above, the C=C section is planar, so these carbons are open on the sides for attack from a halogen.  A bond still has to break for a new C-Cl bond to form, but in this case it is the weaker C-C pi bond that breaks.  With easy physical access and a lower energy barrier, the reaction goes forward.