The trace $\text{Tr}_{F/K}(\alpha)$ of $\alpha$ over $K$ is defined by $$\text{Tr}_{F/K}(\alpha) = \alpha + \alpha^q + \cdots + \alpha^{q^{m-1}},$$ where $\alpha \in F = \mathbb{F}_{q^m}$ and $K=\mathbb{F}_q$. We know that $\text{Tr}_{F/K}$ is a linear transformation from $F$ onto $K$.
Consider the following theorem.
$\textbf{Theorem.}~$ Let $F$ be a finite extension of the finite field $K$, both considered as vector spaces over $K$. Then the linear transformations from $F$ into $K$ are exactly the mappings $L_\beta$, $\beta \in F$, where $L_{\beta}(\alpha)=\text{Tr}_{F/K}(\beta\alpha)$ for all $\alpha \in F$. Futhermore, we have $L_{\beta} \ne L_{\gamma}$ whenever $\beta$ and $\gamma$ are distinct elements of $F$.
$~\\$ For $\beta \ne \gamma$, it is easy to check that $L_{\beta}$ and $L_{\gamma}$ are different. If $K = \mathbb{F}_q$ and $F = \mathbb{F}_{q^m}$, then the mappings $L_{\beta}$ yield $q^m$ different linear transformations from $F$ into $K$. But why the mappings $L_\beta$ exhaust all possible linear transformations from $F$ into $K$?
The explanation from the textbook is that every linear transformation from $F$ into $K$ can be obtained by assigning arbitrary elements of $K$ to the $m$ elements of a given basis of $F$ over $K$. Since this can be done in $q^m$ different ways, the mappings $L_{\beta}$ already exhaust all possible linear transformations from $F$ into $K$.
I am very confused about this explanation. Why the mappings $L_\beta$ exhaust all possible linear transformations from $F$ into $K$? The number of the linear transformations from $F$ into $K$ is always $q^m$?
Since $F$ is an $m$-dimensional vector space over $K$, the $K$-linear maps $F \to K$ is ${\rm Hom}_K(F,K) \cong {\rm Hom}_K(K^m,K)$, which has $K$-dimension $m$ and thus size $|K|^m = q^m$.
But the fact that all $K$-linear maps $F \to K$ have the form you describe in terms of the trace map is not a special property of finite fields: it holds for all finite separable field extensions $F/K$ (they can be infinite fields): each $K$-linear map $F \to K$ has the form $\varphi_\beta : \alpha \mapsto {\rm Tr}_{F/K}(\beta\alpha)$ for a unique $\beta \in F$. That is, the mapping $F \to {\rm Hom}_K(F,K)$ by $\beta \mapsto \varphi_\beta$ is an isomorphism of $K$-vector spaces when $F/K$ is separable. (This is all wrong when $F/K$ is inseparable, since in that case, and only in that case, the trace mapping ${\rm Tr}_{F/K}$ is identically zero.)
By the way, I don't advise saying the trace mapping on finite fields is defined as that sum you use, because there is a trace mapping on finite extensions of all fields even though the description you use for the trace mapping is not valid when the fields are infinite it's better to use a uniform definition of ${\rm Tr}_{F/K}(\beta)$, as the trace of the linear map $L_\beta$. See here.