So Cohn uses the notation that many have wanted to change to, being $xfg$ rather than $g(f(x))$, and I have had the example:
Let $f,g: \mathbb{N} \to \mathbb{N}$, be given by $xf = x + 1,xg=x^2$, then $xfg = (x+1)^2 , xgf=x^2+1$
We can clearly see that $x$ gets put into $f$ which all gets put into $g$ for $xfg$
$xf=x+1,(xf)g=xfg=(x+1)g=(x+1)^2$
But after this we do mappings and are told: (For $S=\{1,2,3\}$)
$f=\begin{pmatrix} 1&2&3\\2&3&1 \end{pmatrix},g=\begin{pmatrix} 1&2&3\\1&3&2 \end{pmatrix}$ which would make me think that $fg=\begin{pmatrix} 1&2&3\\2&1&3 \end{pmatrix}$
But instead $fg=\begin{pmatrix} 1&2&3\\3&2&1 \end{pmatrix}$ which is what I would expect $gf$ to equal and $gf$ equals what I would expect $fg$ to equal.
Why are we doing $xfg$ from left to right, but mapping permutation matrices from right to left?
You have misunderstood how permutations work:
If we have the following, from the set $S=\{1,2,3,4\}$
$f=\begin{pmatrix}1&2&3&4\\3&4&1&2\end{pmatrix}, g= \begin{pmatrix}1&2&3&4\\2&1&3&4\end{pmatrix}$
Then $fg=\begin{pmatrix}1&2&3&4\\3&4&2&1\end{pmatrix}$, since we take the elements and permute them as follows:$$(1\to3\to3)(2\to4\to4)(3\to1\to2)(4\to2\to1)$$
Hence yes, it is consistent notation, meaning it does work from left to right.