One common definition of $[L:K]$ is $\dim_K(L)$, the dimension of $L$ as a $K$-vector space. If that's the definition you're using, then the answer to your first question is
1) $[L : K] = r$ means by definition that $L \cong K^r$, since every vector space of dimension $r$ over $K$ is isomorphic to $K^r$ (in a highly non-unique way!).
If $L$ is a principal extension, meaning that $L = K(\alpha)$ for some single element $\alpha \in L$, then this is the same as the degree of the minimal polynomial of $\alpha$ over $K$.
In our case, since finite fields are separable extensions of the prime field $K = \mathbb{F}_p$, the primitive element theorem tells us that $L \cong K(\alpha)$ for some $\alpha$. Let
$$ m(x) = x^r + a_1x^{r-1} + \cdots + a_r $$
be the minimal polynomial of $\alpha$ over $K$. Then I claim that a canonical basis for $L$ over $K$ is given by the elements $1, \alpha, \alpha^2, \dots, \alpha^{r-1}$. Indeed, we know that these powers of $\alpha$ are linearly independent over $K$, since an equation of linear dependence (supposing the coefficient on $\alpha^{r-1}$ is nonzero, and so can be taken to be 1)
$$ \alpha^{r-1} + b_2 \alpha^{r-2} + \cdots + b_{r-1} = 0 \quad \text{ with $b_i \in K$}$$
gives a polynomial $P(x)$ (by replacing $\alpha$ with $x$) of smaller degree than $m(x)$ such that $P(\alpha) = 0$, contradicting that this was the minimal polynomial. On the other hand, since $m(\alpha) = 0$ we can re-write any polynomial in $\alpha$ of degree $\ge r$ in terms of $1, \dots, \alpha^{r-1}$ using the relation
$$ \alpha^r = -(a_1 \alpha^{r-1} + \dots + a_r), $$
coming from $m(\alpha) = 0$.
We conclude that $1, \alpha, \dots, \alpha^{r-1}$ is a basis for $L$ over $K$ of size $r$.
2) $K^r$ is a vector space over $K$.
3) Yes.
Let me know if what I wrote is still confusing. I'm afraid I don't have a copy of Artin's Algebra nearby, so I can't check what the definition of $[L : K]$ he's using is.