The permutations in your question are given in Cauchy's two-line notation, where the upper line gives the input index to the permutation function, and the lower line gives the resulting permuted index.
For example, the definition $$\sigma = {1\ 2\ 3 \choose 3\ 1\ 2}$$ means the same as $$\sigma(1) = 3,\quad \sigma(2) = 1,\quad \sigma(3) = 2.$$ Thus, if we apply this permutation $\sigma$ to a tuple $x = (x_1, x_2, x_3)$, we obtain the permuted tuple $$x_\sigma = \sigma \cdot x = (x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}) = (x_3, x_1, x_2).$$
In particular, if $x = (x_1, x_2, x_3) = (\text{V}, \text{E}, \text{N})$, then $x_\sigma = (x_3, x_1, x_2) = (\text{N}, \text{V}, \text{E})$.
This is easy visualize if you simply replace the indices $1$, $2$ and $3$ in the definition of the permutation with the elements $x_1 = \text{V}$, $x_2 = \text{E}$ and $x_3 = \text{N}$ to be permuted, giving $${x_1\ x_2\ x_3 \choose x_3\ x_1\ x_2} = {\text{V E N} \choose \text{N V E} }.$$
The second example in your question appears to be incorrect: given the permutation $$\sigma' = {1\ 2\ 3 \choose 2\ 3\ 1}$$ and the input $x' = (x'_1, x'_2, x'_3) = (\text{T}, \text{E}, \text{N})$, the output should be $$x'_{\sigma'} = \sigma' \cdot x' = (x'_{\sigma'(1)}, x'_{\sigma'(2)}, x'_{\sigma'(3)}) = (x'_2, x'_3, x'_1) = (\text{E}, \text{N}, \text{T}),$$ not $(x'_3, x'_1, x'_2) = (\text{N}, \text{T}, \text{E})$. It would, however, be correct if we replaced the permutation $\sigma'$ with its inverse $$\sigma'^{-1} = {2\ 3\ 1 \choose 1\ 2\ 3} = {1\ 2\ 3 \choose 3\ 1\ 2}$$ (which, as it happens, equals the first permutation $\sigma$) obtained by swapping the rows and (optionally) re-sorting the columns in the two-line definition of $\sigma'$.