The following questions are taken from Module Theory by Blyth:
For each $r\in R$, the homothety on $M$ by $r$ is the group homomorphism $$\mu_r:M\to M:x\mapsto r\cdot x,$$ which is $R$-linear if $R$ is commutative. We then define $$H_R(M):=\{\mu_r\mid r\in R\}$$ to be the collection of homotheties on $M$. In fact, the structure ring homomorphism $\Phi:R\to\operatorname{End}(M)$ on $M$ has kernel equal to $\operatorname{Ann}_R(M)$ and image equal to $H_R(M)$, so $H_R(M)$ is a ring satisfying $$R/\operatorname{Ann}_R(M)\cong H_R(M).$$ That is precisely 13.1 shown above, which is an easy exercise.
The most annoying part is 13.2. Here we assume that $Rx$ is simple for all nonzero $x\in M$. To prove the given statement, it suffices to consider the case when $M$ is not simple. Following the hint, we can find a nonzero $x\in M$ such that $M\ne Rx$ and fix an arbitrary $y\notin Rx$. In this case, one can show that $$Ry\cap Rx=0.$$
Assume, to the contrary, that there exists a nonzero element $z\in Ry\cap Rx$. By assumption, now $Ry$ is simple. Since $z\in Ry$ is nonzero, we should have $Ry=Rz$. However, as $z\in Rx$ as well, it follows that $Ry=Rz\le Rx$ whence $y\in Rx$, a contradiction.
Correspondingly, what I could prove was indeed $$\operatorname{Ann}_R(x+y)=\operatorname{Ann}_R(x)\cap\operatorname{Ann}_R(y).$$
The direction $\operatorname{Ann}_R(x)\cap\operatorname{Ann}_R(y)\subseteq\operatorname{Ann}_R(x+y)$ is trivial. Conversely, let $r\in\operatorname{Ann}_R(x+y)$. Then $$0=r\cdot(x+y)=r\cdot x+r\cdot y,$$ implying that $r\cdot x=-r\cdot y\in Rx\cap Ry$. As shown above, we must have $r\cdot x=r\cdot y=0$, so $r\in\operatorname{Ann}_R(x)\cap\operatorname{Ann}_R(y)$.
I am curious about how the author managed to get $+$ rather than $\cap$.
In addition, assume that the hint is true. Now because $Rx$ is simple, one can see that $\operatorname{Ann}_R(x)$ is a maximal left ideal of $R$. Then one must have $\operatorname{Ann}_R(x+y)=\operatorname{Ann}_R(x)$, hence $\operatorname{Ann}_R(y)\le\operatorname{Ann}_R(x)$. By symmetric arguments, one should also have $\operatorname{Ann}_R(x)\le\operatorname{Ann}_R(y)$, so indeed $\operatorname{Ann}_R(x)=\operatorname{Ann}_R(y)$. Then how would people deduce that $\operatorname{Ann}_R(M)=\operatorname{Ann}_R(x)$ from such identity? Of course, once such identity holds, we can see that $\operatorname{Ann}_R(M)$ is a maximal left ideal of $R$. Since it is indeed two-sided, it is a maximal ideal of $R$ whence $R/\operatorname{Ann}_R(M)\cong H_R(M)$ is a division ring.
Therefore, my question consists of two parts:
Is this given hint $\operatorname{Ann}_R(x+y)=\operatorname{Ann}_R(x)+\operatorname{Ann}_R(y)$ correct?
If it is, how would people deduce that $\operatorname{Ann}_R(M)=\operatorname{Ann}_R(x)$?
Any help will be appreciated.