A small note: for the question considered here, the relevant vector space axiom is:
$\alpha \cdot(\beta \cdot v) = (\alpha\beta)\cdot v$ which tells us the group $F^{\ast}$ acts (as a left action) on the set $V$.
It is natural to ask, why can't we define a right action by:
$v \cdot \alpha = \alpha \cdot v$?
Well, we can, but we would like a right-action to satisfy:
$(v\cdot \beta)\cdot \alpha = v\cdot(\beta\alpha) = (\beta\alpha)\cdot v$, and what we get is:
$(v\cdot \beta)\cdot \alpha = (\alpha\beta)\cdot v$.
In a field, or even a commutative ring, this is not an issue, but if one's "action upon an abelian group" is via a non-commutative ring it becomes a problem.
In other words, any left-vector space can be turned into an isomorphic right-vector space over the same field- in the case where we have an extension field of a smaller field, these two are the same entity.
This doesn't contradict alex.jordan's example, as the right-action he exhibits is a distinct action from the "natural" one induced here.
So, for example, when speaking of $R$-modules, if $R$ is commutative, it is typical to "pick a side and stick with it", although there are times (such as when dealing with tensor products) when it is convenient to view an $R$-module, as an $R,R$-bimodule (for a commutative ring $R$).
With vector spaces $R = F$, any vector space is already an $F,F$-bimodule (using the natural "left=right" action), but as alex.jordan's example shows, not ALL $F,F$-bimodule structures arise in this way.