Determine which orbitals will form hybrids with one another
Let's take your questions in reverse order.


  what is the rule in general to determine which orbitals will form
  hybrids?


The "general" rule is that you mix one atomic orbital for each bond you want to form.  In methane, carbon has 4 valence electrons and is forming 4 bonds so it mixes 4 atomic orbitals (1 2s and 3 2p) generating 4 $\ce{sp^3}$ hybrid orbitals, and it places 1 electron in each orbital.  

In the case of aluminum trihydride, aluminum has 3 valence electrons and will form 3 bonds, one to each of the 3 hydrogen atoms.  Therefore, aluminum must mix 3 atomic orbitals (1 3s and 2 3p) generating 3 $\ce{sp^2}$ orbitals and places 1 of its electrons in each of these orbitals.


  why do the lower orbitals form a hybrid?


A 2s orbital is relatively close in energy to a 2p orbital and the same applies for a 3s and 3p orbital.  The closer in energy 2 orbitals are, the more effective is their mixing.  Therefore, 3s and 3p orbitals can mix (form new hybrid orbitals) effectively.  The reason orbitals mix in the first place is to form new orbitals that have greater directionality and can therefore form stronger bonds.  After all, we could form 3 bonds to aluminum trihydride by just bonding to the 3s orbital and separately to 2 3p orbitals.  Two of these bonds would be oriented 90 degrees from one another and the third bond would be fluxional, moving around the surface of a (3s) sphere. But hybridizing the aluminum so that we can form 3 $\ce{sp^2}$ orbitals and use them for bonding with hydrogen produces a more stable molecule and one with the observed geometry (3 bonds in a plane with a 120 degree angle between them).  If hybridization doesn't produce stronger bonds and a molecule that is more stable, then the molecule wouldn't hybridize. $\ce{PH3}$ is an example of such a molecule, the central phosphorous atom remains unhybridized.