Before arising my question, let us see the following fact.
Let $R$ be a ring with unity and $R'$ be a subring of $R$. If $M'$ is a simple $R'$-module, there exists a simple $R$-module $M$ such that $\mathrm{Hom}_{R'}(M',M)\neq\{0\}$.
Proof: By Frobenius reciprocity, it is equivalent to show $\mathrm{Hom}_R(R\bigotimes_{R'}M',M)\neq\{0\}$ for some simple $R$-module $M$. Take any nonzero element $m'\in M'$, and one obtains $M'=R'm'$ because $M'$ is a simple $R'$-module. It follows that $R\bigotimes_{R'}M'=R(1\otimes m')\cong R/\mathrm{Ann}_R(1\otimes m')$, where $\mathrm{Ann}_R(1\otimes m')$ denotes the left ideal of $R$ of the annihilators of $1\otimes m'$. Since $R$ is with unity, one may take a maximal left ideal $I$ containing $\mathrm{Ann}_R(1\otimes m')$. Set $M=R/I$, and $M$ is a simple $R$-module and $\mathrm{Hom}_R(R\bigotimes_{R'}M',M)\cong\mathrm{Hom}_R(R/\mathrm{Ann}_R(1\otimes m'),R/I)\neq\{0\}$.
Now let me raise the following question.
Let $R$ be a ring with unity, $R'$ be a subring of $R$, and $R''$ be a subring of $R'$. If $M$ is a simple $R$-module, and if there exists a simple $R''$-module $M''$ such that $\mathrm{Hom}_{R''}(M'',M)\neq{0}$. Does there always exist a simple $R'$-module $M'$ such that $\mathrm{Hom}_{R''}(M'',M')\neq{0}$ and $\mathrm{Hom}_{R'}(M',M)\neq{0}$?
By the above fact, we may take a simple $R'$-module $M'$ such that $\mathrm{Hom}_{R''}(M'',M')\neq{0}$. But how to adjust this $M'$ such that $\mathrm{Hom}_{R'}(M',M)\neq{0}$?
Let $k$ be a field, and take $R=k(t)$, $R'=k[t]$, $R''=k$, $M=k(t)$, and $M''=k$.
Then every non-zero element of $M$ generates an $R'$-submodule of $M$ isomorphic to $R'$, which is not simple. So there is no simple $R'$-module $M'$ with $\operatorname{Hom}_{R'}M',M)\neq0$.