I found similar threads here but I didn't find any satisfactory answers.
So if $f$ is monic, then given the kernel $(K, k)$ we have $fk = 0 = f0$ and because $f$ is monic we get $k = 0$ and if $K \neq 0$ then there is not a unique morphism from all candidates $(K', 0)$ to $K$ as required by the definition of the kernel so $\ker(f)$ must be $(0, 0)$.
If $\ker(f) = (0, 0)$, then I don't find it as easy. One can show that if $fg = 0$ then by definition we have a unique morphism $l$ s.t. $g = 0l = 0$ so $fg = 0 \implies g = 0$ for all $g$.
However, if we have $fg_1 = fg_2 \neq 0$, then I'm not sure how to prove that it holds. Is there some other condition that we need or am I missing something? Does it work if the category is abelian?
Suppose $fg = fh$, then $f(g-h) = 0$ so $g-h$ factors through $\ker(f) = 0$. Therefore...