Let's begin with two opposite charges, $+q$ and $-q$ placed at a short distance form each other. I assume you know that in the potential they generate the term $1/r$ (the "monopole") is lacking whereas the "dipole" $1/r^2$ term is present. If positions of charges are $$+q:\quad (d/2, 0, 0) \qquad -q:\quad (-d/2, 0, 0)$$ then the dipole moment is vector
$$(qd, 0, 0).$$
To get a pure dipole potential you must take the limit $q\to\infty$, $d\to0$, with $q\,d$ constant.
Now consider 4 charges, $\#1$ positive $+q$, $\#2$ negative $-q$, $\#3$ still $-q$, $\#4$ equal $+q$ again. Let's place them as follows:
\#1:& +q \quad (d/2, d/2, 0) \cr
\#2:& -q \quad (d/2, -d/2, 0) \cr
\#3:& -q \quad (-d/2, d/2, 0) \cr
\#4:& +q \quad (-d/2, -d/2, 0).\cr}$$
You can see that these are two dipoles, of opposite moments, one near the other. Now potential lacks the $1/r^2$ term too; its series begins with $1/r^3$. Needless to say, this is a quadrupole. In order to get a pure quadrupole a limit is to be taken again: $q\to\infty$, $d\to0$, $q\,d^2$ constant.
In general to jump from a multipole to the following one we must pair a multipole with another exactly alike, but for charge signs. So the number of charges gets doubled.
But quadrupole moment is a more complex entity: not a vector but a tensor. Actually a quadrupole may be built in other ways. Just an example.
\#1:& +q \quad (0, 0, d) \cr
\#2:& -2q \quad (0, 0, 0) \cr
\#3:& +q \quad (0, 0, -d).\cr}$$
Its moment is a different tensor and can't be transformed into the former even if you subject it to a rotation.
The instance above is only an apparent exception to the "doubling" rule, as $+q, -2q, +q$ is just two opposite dipoles aligned.
If you've got the game, I can save to show how an octupole is built.