In Wikipedia it says a group $G$is solvable if it has a subnormal series
$\{e\}=G_1\lhd G_2,\dots \lhd G_n=G$ where $G_i$ is a normal subgroup of $G_{i+1}$ and all the factor groups are abelian.
My question is does this only mean the quotient groups $G_i/G_{i-1}$ or the quotients between any pair of Groups in the sequence?
As I mentioned in a comment, only those quotients where the indices differ by $1$ are required to be abelian, as otherwise, this would require that the group itself was abelian.
Here is a good way to think of solvability:
If the group has a composition series (so for example if it is finite), this is a subnormal series where all the quotients (again, with indices differing by $1$) are simple. Now, if the group is solvable, one can check (and it is a nice exercise to do this) that the quotients in the composition series are cyclic of prime order.
So where a general group is "build from" simple groups, a solvable group is "build from" finite groups of prime order.
Another thing to think about is the requirement that each $G_i$ be normal in $G_{i+1}$ (rather than being normal in $G$). Another good exercise is to show that if $G$ is solvable then there is in fact a normal series (i.e. where each $G_i$ is indeed normal in $G$) with abelian quotients (hint: Consider commutator subgroups).
If one takes a normal series which is as "fine" as possible (so we cannot put further subgroups in between the chosen ones and still get a normal series), then one can check (somewhat harder exercise) that the quotients are elementary abelian, so they are in fact vector spaces over a prime field, and this has a lot of important consequences when one studies representations.