Let $q=3$ and $m=4$, and the map $$T(a) = a + a^q + a^{q^2} + \cdots + a^{q^{m-1}}$$
We write $F$ the finite field with $q^m$ elements and $K$ the field with $q$ elements.
Proof that $T(\alpha+\beta) = T(\alpha)+T(\beta)$ if $\alpha,\beta \in F$.
I started re-writing $T(a)=a+a^3+a^9+a^{27}$.
Now, $T(\alpha+\beta)=(\alpha+\beta)+(\alpha+\beta)^3+(\alpha+\beta)^9+(\alpha+\beta)^{27}$.
So $(\alpha+\beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3$. I need the first and last term to make equality occurs, but how can I delete the second and third term? $F$ is a field with $81$ elements, I cannot suppress the terms because ther are $3\cdot t$.
How can I proceed?
There is only one field of order 81 (up to isomorphism). Therefore your $F$ is isomorphic to $\mathbb{F}_{81}$, which is of characteristic $3$. Thus $(a + b)^3 = a^3 + b^3$ holds and you can suppress the terms you want.