Why is it a consequence of a notational choice that left actions are covariant functors? Could someone elaborate on this remark? I don't see what the notation has to do with it.

That left actions are covariant functors and right actions are contravariant
  functors is a consequence of a basic notational choice: we write the value of a
  function $f$ at an element $x$ as $f(x)$, not $(x)f$.
  (https://arxiv.org/pdf/1612.09375.pdf; 1.2.14)

 A: Our notational choice regarding functions affects what order we apply functions in a function composition ($f\circ g$ means "first apply $g$, then apply $f$", because that's the natural interpretation when you write $f\circ g(x)$), which in turn is what affects whether a covariant functor yields a left or right action.
Let $G$ be a monoid, and let $\mathscr G$ be that monoid realized as a category with a single element $x$. The elements of $G$ correspond to morphisms $x\to x$.
Say we write function application the regular way (i.e. $f(x)$). Take a covariant functor $F: \mathscr G\to\mathbf{Set}$. This induces an action of $G$ on the set $S=F(x)$ by way of either $g\cdot s=F(g)(s)$, or $s\cdot g=F(g)(s)$. Which one is correct?
If we take two elements $g,h\in G$, then $gh$ is represented in $\mathscr G$ by the morphism $g\circ h$. Applying $F$ gives us $F(g\circ h)=F(g)\circ F(h)$. This is a function on $S$, and the way we write functions, it means we first apply $F(h)$, then $F(g)$. In other words, $(gh)\cdot s=g\cdot (h\cdot s)$, not $s\cdot (gh)=(s\cdot g)\cdot h$.
If $F$ were contravariant, then $F(g\circ h)=F(h)\circ F(g)$, which by a similar argument gives a right action.
Now what happens if we were to change notational convention? Then $(g)F\circ (h)F$ would mean "First apply $(g)F$, then apply $(h)F$", which would completely reverse the above argument. A covariant functor would still conserve the order of a composition, while a contravariant functor would reverse it, but this time the covariant functor is the one that interprets $gh\in G$ as applying $(g)F$ first.
