Suppose $R$ is a ring and $M,N$ are $R$-modules and $f:M\to N$ is an $R$-linear map.
If $\frak{p}$ is a prime ideal of $R$, then we have a $R$-linear map $f_{\frak{p}}:M_{\frak{p}}\to N_{\frak{p}}$ (where $M_{\frak{p}}:=M/\mathfrak{p}M$ and $N_{\frak{p}}:=N/\mathfrak{p}N$) defined by $f_{\frak{p}}([m])=[f(m)]$.
I'm trying to prove that $(\ker (f))_{\frak{p}}=\ker (f_{\frak{p}})$.
It is obvious to check inclusion $\subset$, but I'm having trouble with $\supset$. Suppose $[m]\in\ker(f_{\frak{p}})$. Then, $[0]=f_{\frak{p}}([m])=[f(m)]$, so $f(m)\in\mathfrak{p}N$, or $m\in f^{-1}(\mathfrak{p}N)$. I'm trying to prove that $f^{-1}(\mathfrak{p}N)=\mathfrak{p}M+\ker(f)$, which seems pretty intuitive, but I don't know how to actually prove it.
If you really want $M_{\mathfrak{p}}=M/{\mathfrak{p}}M$ then you are out of luck. This is tensoring by $R/\mathfrak{p}$ which is not a left-exact functor. As a counterexample, take $R=\mathbb{Z}$, $M=N=\mathbb{Z}$, $\mathfrak{p}=2\mathbb{Z}$ and $f:n\mapsto2n$.
But is it more usual for $M_\mathfrak{p}$ to denote the localisation of $M$ at $\mathfrak{p}$ (the set of "fractions" $m/a$ where $m\in M$ and $a\in R-\mathfrak{p}$ under a suitable equivalence). This operation is left-exact: it does preserve kernels as you want. Is this really what you mean?