Localisation of modules is localisation of category

Question

I have spent a few days trying to solve the following problem and can't move forward in it, would appreciate any hint:
Let $S$ be a multiplication closed subset of some commutative ring $R$. Let $Q$ be a collection of morphisms every element of which kernels and cokernels is annulled by multiplication on some element of S. Localisation is defined as pair (category $J = R-mod[Q^{-1}]$, functor $j$) that correspond to the following property: (1)$j$ sends elements of $Q$ to isomorphisms and (2)they satisfy an universal property: for every pair (category $Y$, functor $y$) that satisfy (1) there is unique functor from $J$ to $Y$ that force appearing diagram to commute. We can also define a functor of localisation of module that sends module $M$ to element $(M, S^{-1}M)$ of category $Z$ consisting of pairs (module, it's localisation) with morphisms borrowed from morphisms between localisations.
I have chosen a path of proving universal property of localization of category for pair $(Z, z)$ where z is functor sending $M$ to $(M, S^{-1}M)$ and $\square\overset{\phi}{\to}\square$ to corresponding morphism of localisations where it's enough to define it on elements like $\frac{m}{1}$. While moving in circles around this problem i noted the following: there are morphisms that seem to form some kind of base of $Q$: $\psi_{s}: m\mapsto s_{m}m, s_m\in S$. Their kernels' and cokernels' localization is zero and since $z$ preserves exactness of sequence of two morphisms it sends $\psi_{s}$ to isomorphism due to decomposition $0\longrightarrow ker \longrightarrow dom\longrightarrow image\longrightarrow codom \longrightarrow coker \longrightarrow 0$ going to $0 \longrightarrow dom\longrightarrow image\longrightarrow codom \longrightarrow 0$. In result $z(\psi_{s})$ has a reverse in category $Z$. Then, remembering the explicit construction of localisation of category, i was trying to decompose any morphism in $Z$ into composition of some morpisms that have preimages in $C$ and new ones: dividing elements, but had no luck at it.