First, CH3 group is not a leaving group so it can't be displaced. Only the chloride can leave.
Even if CH3 could leave, there would be more steric hindrance than on the other end. CH3 is a larger group than Cl. So it would still react this way. (But it's a moot point since CH3 is not LG.)
E2 is a concerted reaction thus it does not go rearrangements. It still prefers forming a double bond towards more substituted side unless there is a bulky base present
my understanding was, was that a reaction would undergo E2 with a bulky nucleophile, or with an alkoxide (given that the leaving group is on a tertiary or secondary carbon)
No the OP is right. He showed the Cl leaving. The problem with the reagents is though that you are using a very strong base in this case, NaOH. Once chlorine decides to leave, OH sees its moment of opportunity and attacks one of the hydrides before the molecule would have time for rearrangement. Usually with strong bases you see E2 without the formation of a carbocation. You would want to use a weaker base to go through with E1. Hopefully that clears it up for you.
Rearrangement can only happen because a carbocation is formed.
In E2 reaction, since the reaction is concerted the carbocation never forms.
If it is a strong base, E2/ will happen. A strong base AND a strong nucleophile will do E2 and SN2 (favors one or the other depending on the electrophile)
E1 generally doesn't happen often. It is usually a side-product of SN1 reaction.
Maybe with a weak base/nucleophile and a tertiary carbon with heat.