Let $R$ be a (not necessarily commutative) ring with unity and $M_R$ be a right $R$-module. Define the right (resp. left) annihilator of $a\in R$ as $\text{Ann}^{R}_r(a)=\{r\in R:ar=0\}~(\text{resp.}~~\text{Ann}^{R}_l(a)=\{r\in R:ra=0\})$.
Why for modules I don't see any emphasis put on the side of the annihilators (left or right)? They are normally written as $\text{Ann}^{M}(a)=\{m\in M:ma=0\}$. I even encountered the following:
For an element $x$ in $R$, if $Max=0$, then $Mx\subseteq \text{Ann}^{M}(a)$. This is where I have a problem. I feel it should be that $Ma\subseteq \text{Ann}^{M}(x)$ because $(Ma)x=0$. I don't clearly see how $(Ma)x=0$ becomes $(Ma)(Mx)=0$?
Your first question: there's no need to specify the side for $\operatorname{Ann}^{M}(a)$ because we already know the side. In the given context, knowing that $M=M_R$ is a right $R$-module, we know that between an element $m\in M$ from the module $M$ and an element $a\in R$ from the ring $R$, the only meaningful multiplication is $ma$, but not "$am$".
As for the second part of your question, I'm also sure that this is a typo. Maybe they meant to say that "if $Max=0$, then $Ma\subseteq\operatorname{Ann}^M(x)$"; or maybe they meant to say that "if $Mxa=0$, then $Mx\subseteq\operatorname{Ann}^M(a)$". (These are really the same statement, up to renaming.) But, to be honest, your reasoning in the end doesn't make sense: the expression "$(Ma)(Mx)$" is not meaningful, because we can't multiply elements of a module with each other.