Maybe I'm underestimating your problem, but it seems Mikael above is right.
In your example you define $q:f_1\to f_2$, so if it's a kernel of some other map r, then $r$ must have $f_2$ for domain. $q$ can't possibly be a kernel of $p$, as the composition $pq$ does not make sense.
Given a map $f: X \to Y$, a kernel is another map $k:K \to X$ satisfying blah blah.
Now, if X and Y are complexes you have a criterion to check wether $k$ is a kernel: checking the components $k^n$ (but you already must have a chain map $k$ to begin with).