It's difficult to tell from the way you have represented it in the last diagram. I take it that the straight line, as indicated from the first two diagrams, is meant to indicate the R group. However, in the last diagram, you seem to have placed the R group in between the second carbon and a CH3 group, which would make the CH3 group not part of that chain. I'm sure this isn't what you meant to do, and the CH3 is meant to be part of the chain. Could you find a way to make it more clear where the R group is actually attaching?
Keep in mind that the iso- term, if I'm not mistaken, generally is used for groups that arranged in a symetrical fashion. So Isopropyl (3 carbons) would be (with the dots used for the same purpose as yours, space-holders for SDN)
....CH3CHCH3
.........*
with the asterik meant to represent to bond between the second carbon and the R group. So this is not sec-propyl, but rather isopropyl because the alkyl group is symetrical. A similar scheme works to make isobutyl:
CH3CHCH3
.....CH2
.....*
wherein the 4 C's form a sort of "T", and the tip of the T is bonded (*) to the R group. Again, the alkyl group is now symmetrical, and thus the term isobutane.
Be careful, because if the alkyl is constructed like this:
.....*
CH3CCH3
.....CH3
with the R group bonded to the second C in the top part of the "T" it is called tert-butyl. It is called tert-butyl because though it has a "T" shape like isobutyl, in this case the R group is bonded to a tertiary carbon.
That should help you out, though I would DOUBLECHECK this. This is off the top of my head and I'm kind of tired. If I have made a mistake, it was in describing one as tert and the other as iso; if I'm wrong it is simply the other way around and you can still compare it to the isopropyl.