Consider two sequences $A\overset{f}\to B\to C\to0$ and $0\to A\to B\overset g\to C$ of abelian groups.
The first one is exact iff $C\cong\operatorname{coker}f$ as a consequence of the first isomorphism theorem.
But why do we have that the second is exact iff $A\cong\ker g$ ?
Only one direction is true. Suppose your second sequence is exact. The image of $A \to B$ must be iso to $A$ since that map has trivial kernel. But this image is the kernel of $g$ by exactness. The converse is false.