One can easily see that countability also imposes or enables an ordering of the set.
Now $\mathbb{R}$ is ordered but not countable.
Is this ordering a weaker version of countability (at least in some sense)? If so, what about sets that are both un-countable and not ordered (like $\mathbb{C}$)?
On
You can have an order on $\Bbb C$, just not one that respects the field operations. Lexicographic is a convenient one, or you can choose your favorite bijection with $\Bbb R$ and use that order. You can also have a partial order (or non-strict total order) based on the magnitudes of the points. I don't think I would call it a version of countability.
Every set can be partially ordered. This is easy, just take $\leq_A=\{(a,a)\mid a\in A\}$ and verify that this is a partial order.
Being countable means that you can more than just partially ordered. You can be well-ordered. Namely, a linear ordering (which itself is a stronger requirement than just partial ordering) and that every non-empty set has a least element.
In fact, the real kicker about being countable is that you can well-order the set so that every element has only finitely many predecessors. Something might not be doable otherwise.
On the other hand, $\Bbb R$ is linearly ordered quite naturally. So every set which have the same size as $\Bbb R$ can be linearly ordered. In the case of $\Bbb C$ you can even do that explicitly: $a+bi\leq c+di\iff a<c\lor (a=c\land b\leq d)$ where $a,b,c,d\in\Bbb R$ and $i$ is the square root of $-1$.
The generalization of countability, however, is well-orderability. Meaning that there is a linear ordering of the set that every non-empty subset has a minimum with respect to that order. The axiom of choice guarantees that every set can be well-ordered. But without it it is consistent that there are sets which cannot be well-ordered, for example it is consistent that $\Bbb R$ cannot be well-ordered.
Note, however that such well-ordering need not agree with other naturally occurring structure on the set. So if $\prec$ is a well-ordering on $\Bbb R$, and so $\Bbb R\setminus\{0\}$ has a minimum element with respect to $\prec$, but certainly this element is not the minimum with respect to the natural $<$ we have on $\Bbb R$.