Right module vs left module I would like to have your help on this. Consider the following diagram summarizing the opposition left module-right module:
Left module vs Right module (s an t represent scalars)
Left module:
$$s(x + y) = sx + sy$$
$$(s_1 + s_2)x = s_1x + s_2x$$ 
$$s(tx) = (s t)x$$
Right module 
$$(x + y)t = xt + yt$$
$$x(t_1 + t_2) = xt_1 + xt_2$$
$$(xs)t = x(s t)$$
Well, the thing is, I somewhere found this statement: ".....the distinction is not purely syntactical, since it implies two different associativity rules linking multiplication in a module with multiplication in a ring."
Could anyone be more specific on that and illustrate the diffence so that it becomes clear it is not simply a syntactic one. Would it be a semantic one or what exactly? 

The key problem is, you want $x(nm)=((x)n)m=(nx)m=mnx=x(mn)$.
If you have a ring, $R$, you can define the opposite ring with multiplication flipped (i.e. $x*y=yx$). We can denote this ring $R^{op}$. Now if we have a left $R$-module, $M$, we can define a right $R^{op}$-module structure on $M$, by $x*m=mx$. If can be checked that $(xm)n=nmx=x(m*n)$, and all the others are trivial. This creates a bijection between the categories of left $R$-modules and right $R^{op}$-modules. 
In the commutative case, clearly $R=R^{op}$, so the distinction is meaningless. But there are examples of rings with $R\neq R^{op}$. Sadly there are no simple examples to my knowledge, but some are discussed here: https://mathoverflow.net/questions/64370/simplest-examples-of-rings-that-are-not-isomorphic-to-their-opposites
