Let $D$ be a division ring, and $R=M_n(D)$, the ring of $n\times n$ matrices over $D$. Let $$V=\begin{Bmatrix} \begin{bmatrix} D & 0 & \cdots & 0\\ \vdots & \vdots & \cdots & \vdots\\ D & 0 & \cdots & 0\\\end{bmatrix}\end{Bmatrix}$$ be the subset of $R$ consisting of matrices with $2$nd, $3$rd, .. $n$th columns zero.
Then $V$ is a left simple $R$-module (in fact, $R$ has, up to isomorphism, a unique simple left $R$-module).
Consider $\operatorname{End}(_RV)$, the ring of endomorphisms of the left $R$-module $V$.
In the book A First Course in Non-commutative Rings by Lam, author asserts (see p.33)
(3) The endomorphism ring $\operatorname{End}(_RV)$, viewed as a ring of right operators on $V$, is isomorphic to $D$.
Q. The statement of (3) is concerning isomorphism of rings; but then, what is the need to consider the former ring to be ring of right operators on $V$?
In answer to a similar question see here, it is mentioned that This (writing endomorphisms of left module to right side) avoids awkward references to the opposite ring that pop out when morphisms are written on the same side.
I didn't understand how does, after writing endomorphisms of left module to left side, pops up opposite ring?
I just thought following lines: see if this is correct, which shows some arrival of opposite ring
$V$ is simple left $R$-module, hence it is generated (as $R$-module) by single element, say $v$. Then for any $f\in \operatorname{End}(_RV)$,
(i) $f$ is completely determined by effect on $v$;
(ii) $f(v)=r_f v$ for some $r_f\in R$.
So, if $g$ is another $R$-endomorphism of $V$, then, with convention of writing endomorphisms on left side, we have $$(f g)(v)=f(g(v))=f(r_g v)=r_g(f(v))=r_gr_fv \mbox{ and } (fg)(v)=r_{fg}(v).$$
Since $R$ is simple ring and $V$ is simple $R$-module, so left-annihilator of $v$ is $0$. Thus, from $r_{fg}v=r_gr_f v$ we get $r_{fg}=r_gr_f$. This could be the place of opposite ring to come in picture. Is this right?