From reddit.com
Take any ring $R$, and look at the polynomial ring $R[x]$. For any polynomial $f(x) \in R[x]$, consider the polynomial $F(x,h) = f(x+h)-f(x)$ in $R[x,h]$. This is obviously in the kernel of the quotient map $R[x,h] \to R[x]$, sending $h$ to $0$, which is just the ideal $(h)$. Thus $f(x+h)-f(x) = hG(x,h)$ for some $G(x,h) \in R[x,h]$, or to put it another way, $G(x,h) = \dfrac{f(x+h)-f(x)}h$ is a polynomial in $x$ and $h$.
One can now simply define $f'(x)$ to be $G(x,0)$, which is the same thing as setting $h=0$ in the difference quotient $\dfrac{f(x+h)-f(x)}h$. This gives exactly the same thing as you would get if you took a limit $h\to 0$, but it does it purely algebraically.
So $F$ is a map from $R[x, h] \to R[x]$, and the kernel of $F$ sends $h$ to $0$. So far so good. But where does $G(x, h)$ come from? It's obviously not the image of $R$ (that would be $R[x]$). Is it the Kernel, and is $hG(x, h)$ a co-set of the kernel?
No, $F$ is not a map. $F$ is an element of $R[x,h]$.
$G$ is the unique element of $R[x,h]$ satisfying $F = hG$. The existence of $G$ is guaranteed by the fact that $F$ lies in the kernel of the map $\phi:R[x,h] \rightarrow R[x]$ sending $h$ to $0$, and this kernel is the ideal generated by $h$. The uniqueness is guaranteed by the fact that $h$ is not a zero divisor in $R[x,h]$.