Let's look at an example to understand what $T$ does. Take the case of $q = 3$. The space $V$ is infinite dimensional, and consists of all "formal sums" of the form
$$
p(x) = a_0 + a_1 x + a_2x^2 + \cdots + a_n x^n
$$
where each $a_i$ is taken from $\Bbb F_3$ (whose elements are $0,1,2$).
Now, the transformation $T$ produces the "function" associated with $p$. For example: if $p(x) = 2x^3$, then $Tp$ is the function $f_p:\Bbb F_3 \to \Bbb F_3$ defined by
$$
f_p(0) = p(0) = 2(0)^3 =0, \quad f_p(1) = 2(1)^3 = 2, \quad f_p(2) = 2(2)^3 = 1
$$
Notice that the polynomial $q(x) = 2x$ produces exactly the same function. That is, $p(x) = q(x)$ for every $x \in \Bbb F_3$. However, $p(x)$ and $q(x)$ are distinct as polynomials. In brief: we have $p \neq q$, but $f_p = f_q$. That is, $p \neq q$, but $Tp = Tq$.
Note that $Func(\Bbb F,\Bbb F)$ has a natural choice of basis. Namely, for $k \in \{0,1,\dots,q-1\}$, define $f_k : \Bbb F_p \to \Bbb F_p$ by
$$
f_k(j) = \begin{cases}
1 & j = k\\
0 & j \neq k
\end{cases}
$$
why is this set a basis?
