Let $\mathbb{F}$ be a finite field with $q$ elements and $\mathbb{E}$ an extension field of degree $n$ of $\mathbb{F}$.
How I will be able to demonstrate that $$\sum_{k=0}^{n-1}t_k\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}(g_{ij})^{q^k}x^{q^{i+k}+q^{j+k}}= \sum_{k=0}^{n-1}t_k\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}(g_{i-k,j-k})^{q^k}x^{q^{i}+q^{j}}$$
applying Cyclic Rotation?
$i-k$ and $j-k$ are calculated $\mod n$. I'm trying to use variable change, for example for $i$, $m = i+k \mod n$, but I can't get change the índice sumation $i$
This is just a matter of chopping the sum into two different pieces and reassembling. Given that, for instance, $f(j,k)$ only depends on the value of $j\bmod n$ (i.e., that $f(j+n, k) = f(j,k)$), then we can write: $$\begin{align} \sum_{k=0}^{n-1}\sum_{j=0}^{n-1}f(j+k, k) &= \sum_{k=0}^{n-1}\sum_{m=k}^{n+k-1}f(m,k) &\text{(substituting $m=j+k$)}\\ &= \sum_{k=0}^{n-1}\left(\sum_{m=k}^{n-1}f(m,k)+\sum_{m=n}^{n+k-1}f(m,k)\right)&\text{(splitting at $m=n$)}\\ &=\sum_{k=0}^{n-1}\left(\sum_{m=k}^{n-1}f(m,k)+\sum_{m=0}^{k-1}f(m,k)\right)&\text{(shifting the second sum)}\\ &=\sum_{k=0}^{n-1}\left(\sum_{m=0}^{k-1}f(m,k)+\sum_{m=k}^{n-1}f(m,k)\right)&\text{(swapping the two terms)}\\ &=\sum_{k=0}^{n-1}\sum_{m=0}^{n-1}f(m,k)&\text{(recombining)}\\ \end{align} $$
You can apply this technique to both the sums over $i$ and over $j$ in your equation.