Let $R$ be a commutative ring with unity with and let $I$ be a proper ideal. (I'm not assuming $R$ is Noetherian.) For every $M \in R$-Mod, let $\Gamma_I(M):=\{m \in M : I^n m=0$ for some $n\ge 1\}$.
If $f \in \mathrm{Hom}_R (M,N)$, it can be seen that $f(\Gamma_I(M)) \subseteq \Gamma_I(N)$, giving us a map $\Gamma_I (f):=f|_{\Gamma_I(M)} \in \mathrm{Hom}_R (\Gamma_I(M) , \Gamma_I(N))$. Thus we have a functor $\Gamma_I : R$-Mod $\to R$-Mod.
How to show that $\Gamma_I$ is naturally isomorphic to the functor $\varinjlim \mathrm{Hom}_R(R/I^n, -)$ ?
$\require{AMScd}\newcommand\Hom{\operatorname{Hom}}\newcommand\Im{\operatorname{Im}}$For every $n\in\Bbb N$ we have an injective $R$-module homomorphism: \begin{align} &\tau_M^n:\Hom_R(R/I^n,M)\rightarrowtail\Gamma_I(M)& &\xi\mapsto\xi(1+I^n) \end{align} Then for every $n\leq m$ we have a commutative diagram \begin{CD} \Hom_R(R/I^n,M)@>>>\Hom_R(R/I^m,M)\\ @V\tau^n_M VV@VV\tau^m_MV\\ \Gamma_I(M)@=\Gamma_I(M) \end{CD} hence we get an injective $R$-module homomorphism $$\tau_M:\varinjlim_{n\in\Bbb N}\Hom_R(R/I^n,M)\to\Gamma_I(M)$$ Since $$\Im\tau_M=\bigcup_{n\in\Bbb N}\Im\tau_M^n=\Gamma_I(M)$$ it follows that $\tau_M$ is, in fact, an $R$-module isomorphism. Finally, for every $R$-module homomorphism, $\varphi:M\to N$, the commutative diagram \begin{CD} \varinjlim\Hom_R(R/I^n,M)@>>>\Hom_R(R/I^n,M)@>\tau_M^n>>\Gamma_I(M)\\ @VVV@VVV@VV\Gamma_I(\varphi)V\\ \varinjlim\Hom_R(R/I^n,N)@>>>\Hom_R(R/I^n,N)@>>\tau_N^n>\Gamma_I(N)\\ \end{CD} proves the naturality of $\tau_M$ respect to $M$ giving rise to a natural isomorphism $$\tau:\varinjlim\Hom_R(R/I^n,-)\to\Gamma_I$$