Suppose I have 3 particles a,b,c, a is negatively charged, b is positively charged and c is neutral. We let the charges of a,b be |a|>|b|. If say a were to touch c the neutrally charged particle and then c were to touch b, what happens here?
Elementary particles can't transfer charges by touching.
Let's consider composite particles (like atoms) that can transfer charges, and atom $a$ having greater negative charge than atom $b$ has positive. This is caused by atom $a$ having $n$ more electrons than protons, and atom $b$ having $m$ less electrons than protons, where $n>m$. If the pull from the nucleus of atom $b$ is strong enough, some electrons from atom $a$ will be attracted to atom $b$ strongly enough that they either leave atom $a$ completely, or they create a chemical bond with atom $b$ that will hold together by $a$'s electrons being attracted by $b$'s nucleus, and $b$'s electrons attracted by $a$'s nucleus.
In case of three atoms, $a$ might transfer some electrons to $c$, which then might transfer some to $b$.
In case of macroscopic objects, the charge is caused by charge carriers, such as free electrons (usually), ions (like in electrolytes or plasma) or holes (like in semiconductors).
For example, let's have two pieces of metal $a$ and $b$, $a$ with abundance of free electrons on the surface giving it negative charge, and $b$ with lack of free electrons on the surface giving it positive charge. When $a$ and $b$ are brought into physical contact, some free electrons from $a$'s surface move to $b$'s surface to equalize the electric potential of both $a$ and $b$.
But how much charge is transfer?
For simplicity, let's consider two spherical conductors $a$ and $b$. $a$ has charge $-Q_a$ and $b$ charge $Q_b$, where $Q_a>Q_b$. It can be shown that the electric potential of $a$ is $V_a=k\frac{-Q_a}{r_a}$ and the electric potential of $b$ is $V_b=k\frac{Q_b}{r_b}$, where $r_a$ is the radius of $a$, $r_b$ is the radius of $b$ and $k$ is Coulomb's constant. If we denote the charge moved from $a$ to $b$ by $Q_\Delta$, we get
$$k\frac{-Q_a-Q_\Delta}{r_a}=k\frac{Q_b+Q_\Delta}{r_b}$$
because the new potentials must be equal. By solving for $Q_\Delta$, we get
$$Q_\Delta = -\frac{Q_ar_b + Q_br_a}{r_a + r_b}$$
which is the charge that will move from $a$ to $b$.
