Let $E$ be a finite field of order $p^n$ where $p$ is some prime number. Let $F=\{0,1,2,\cdots,p-1\}$ be the subfield of $E$ generated by $1$. We view $E$ as an $F$-vector space in the following manner:
(1) $E$ is an abelian group under addition.
(2) If $\alpha\in F$ and if $x\in E$, we define $\alpha\cdot x$ to be the product of $\alpha$ and $x$ in the field $E$. Note that this defines "scalar multiplication" of elements of $F$ by elements of $E$.
You should now check that $E$ is an $F$-vector space. For example,
(1) If $\alpha\in F$ and if $x,y\in E$, then $\alpha\cdot(x+y)=\alpha\cdot x + \alpha\cdot y$ since multiplication is distributive over addition in the field $E$.
(2) If $\alpha,\beta\in F$ and if $x\in E$, then $(\alpha+\beta)\cdot x=\alpha\cdot x + \beta\cdot x$ by similar reasoning to that of (1).
(3) If $\alpha,\beta\in F$ and if $x\in E$, then $\alpha\cdot(\beta\cdot x)=(\alpha\beta)\cdot x$, where $\alpha\beta$ is the product of $\alpha$ and $\beta$ in the field $F$. This follows from associativity of the multiplication in $E$.
(4) Finally, $1\cdot x=x$ for all $x\in E$ since the identity $1$ of $F$ is also the identity of $E$.
Therefore, $E$ is indeed an $F$-vector space as claimed.
Vector spaces occur in unexpected ways throughout mathematics and it is therefore important to get used to this particular example. For example, if you have studied the theory of field extensions, you will know that if $F\subseteq E$ are fields, then $E$ can be viewed as an $F$-vector space. A basic (but important) result in this situation is that if $E$ is a finite-dimensional vector space over $F$, then $E$ is algebraic over $F$. (Of course, the converse fails as you should check.)
We know that $GF(q^2)$ is the splitting field of a polynomial of degree $2$ over $GF(q)$. If $\alpha$ is a root of this polynomial in $GF(q^2)$, then $[GF(q)][\alpha]=GF(q^2)$ as you should check. In this case, we know by elementary field theory that $\{1,\alpha\}$ is a basis for $GF(q^2)$ over $GF(q)$. (Intuitively, since $\alpha$ satisfies a polynomial of degree $2$ over $GF(q)$, we can write $\alpha^2$ as a $GF(q)$-linear combination of $1$ and $\alpha$. Repeat this procedure for higher powers of $\alpha$.)