In cracking, why does C form double bonds?
When we heat hydrocarbons to high temperatures (somewhat lower temperatures if we use a catalyst) they crack.  That means that carbon-carbon bonds are broken and smaller fragments are formed.  In the top part of the following drawing, I've picked n-heptane as an example.  You can see that there are 3



different bonds we can break and a number of different fragments will result.  Each of the fragments that result from breaking a C-C bond has an unpaired electron (there were two electrons in the bond we broke, each fragment retains one electron).  These fragments with an unpaired electron are called radicals.  The carbon with the unpaired electron does not have a stable octet configuration, therefor radicals are extremely reactive species and will react rapidly in a variety of ways to regain their stable octet configuration.  

I've picked one of the possible radicals generated in the top drawing and show how it may react further in the bottom half of the figure.  Sometimes it will break a C-H bond and lose a hydrogen atom and form a stable compound with a double bond (an olefin), sometimes it will remove (abstract) a hydrogen from a hydrocarbon (heptane in our example) to generate a new alkane (pentane) and a new radical (perhaps a radical not formed in the initial cracking step).  Many of the radicals formed can rearrange to more stable radicals, this adds further complexity to the overall process.

So starting with just heptane we can get a large variety of saturated hydrocarbons and unsaturated hydrocarbons.  The situation becomes exponentially more complex if, as usual, we do have many more hydrocarbons present in addition to heptane.  

The key here is that radicals are very reactive species - they want to stabilize by regaining the octet of electrons around the carbon atom that has the radical.  They can get that octet and become stable by adding a hydrogen atom (alkane formation) or breaking a C-H bond and eliminating a hydrogen atom to form an alkene.