A priori, if we have an abelian group $V$ (the abelian group structure provides the addition of a vector space), and we give it the structure of a vector space over a field $F$, then we only know how to make $V$ a vector space over $F$, and over any subfield of $F$. This is because when we give $V$ the structure of a vector space over $F$, the information we have specified is how to multiply elements of $V$ by elements of $F$. If $L\subset F$ is a subfield of $F$, then we already know how to define multiplication of elements of $V$ by elements of $L$: elements of $L$ are also elements of $F$, and we just use our definition for them!
For example, the collection of ordered pairs of complex numbers, $V=\mathbb{C}^2$, is an abelian group under the usual addition
$$(\alpha_1,\alpha_2)+(\beta_1,\beta_2)=(\alpha_1+\beta_1,\alpha_2+\beta_2) \text{ for all }(\alpha_1,\alpha_2),(\beta_1,\beta_2)\in V
.$$
It can be given the structure of a vector space over $\mathbb{C}$ by defining
$$\lambda(\alpha_1,\alpha_2)=(\lambda\alpha_1,\lambda\alpha_2)\text{ for all }\lambda\in\mathbb{C},\,\,(\alpha_1,\alpha_2)\in V.$$
But, now that we've done that, it is also a vector space over $\mathbb{R}$, which is a subfield of $\mathbb{C}$ - we know how to multiply elements of $V$ by real numbers because we already have specified how to multiply by complex numbers.
However, the abelian group $V$ cannot be given the structure of a vector space over $\mathbb{Z}/p\mathbb{Z}$ where $p$ is a prime number, which is a field that is not a subfield of $\mathbb{C}$. This is because we would have to have $$p\cdot (\alpha_1,\alpha_2)=(p\alpha_1,p\alpha_2)=0$$ for any $(\alpha_1,\alpha_2)\in V$, which is false.
Finally, I would point out that even if $L$ is not a subfield of $F$, that doesn't prevent $V$ from also being able to be given the structure of a vector space over $L$. In our example of $V=\mathbb{C}^2$, suppose we had originally specified that $V$ was to be considered as a vector space over $\mathbb{R}$. That is, suppose we had said, "Here is our abelian group $V=\mathbb{C}^2$, and we make it into a vector space over $\mathbb{R}$ by defining
$$c(\alpha_1,\alpha_2)=(c\alpha_1,c\alpha_2)\text{ for all }c\in\mathbb{R},\,\,(\alpha_1,\alpha_2)\in V."$$
This wouldn't change the fact that it can also be given the structure of a vector space over $\mathbb{C}$, in a way that agrees with the original structure over $\mathbb{R}$, even though $\mathbb{C}$ is a larger field than $\mathbb{R}$.