$\renewcommand{\phi}{\varphi}\newcommand{\id}{\text{id}}$Consider an endomorphism $\phi:V\to V$, where $V$ is any finite dimensional $F$-vector space. Assume the characteristic polynomial $\chi_\phi=fg\in F[x]$, where $f,g$ are monic polynomials that are relatively prime. Set $W_1=\ker f(\phi)$ and $W_2=\ker g(\phi)$. Show $V=W_1\oplus W_2$.
Here is what I have done:
By Cayley-Hamilton, we have $\chi_\phi(\phi)=0$, thus $f(\phi)g(\phi)=0$. So either $f(\phi)=0$ or $g(\phi)=0$. This means either $\ker f(\phi)=V$ or $\ker g(\phi)=V$, so either $W_1=V$ or $W_2=V$. Thus $V=W_1+W_2$. We now show the sum is direct. From our previous work we know either $f(\phi)=0$ or $g(\phi)=0$, so assume without loss of generality that $f(\phi)=0$. Since $f$ and $g$ are relatively prime, there are polynomials $h_1,h_2\in F[x]$ such that $h_1f+h_2g=1$. Thus $h_1(\phi)f(\phi)+h_2(\phi)g(\phi)=\id_V$. But $f(\phi)=0$, so $h_2(\phi)g(\phi)=\id_V$. For any non-zero $x\in V$, we have $h_2(\phi(x))g(\phi(x))=x$, so $g(\phi(x))$ cannot be zero. Thus $\ker g(\phi)=W_2$ is trivial, so $W_1\cap W_2=0$, hence $V=W_1\oplus W_2$.
I am wondering if this line of reasoning holds, since I feel a bit weird about the fact that one of the summands ends up being the whole space and the other one ends up being empty. I also don't think I ever used the assumption that both polynomials are monic, but does this even matter?
Since $f(\varphi)$ and $g(\varphi)$ are endomorphisms of $V$, their product can be zero even if none of them are (e.g. two orthogonal projections to orthogonal complementary subspaces).
Restart the proof, based on your polynomials $h_1,h_2$ given by the Bezout identity, and its consequence: $$h_1(\varphi)f(\varphi)+h_2(\varphi)g(\varphi)=\mathrm{id}_V\,.$$
And indeed, the assumption that $f$ and $g$ are monic is not important here. Nevertheless, their product is monic, since it's assumed to be the characteristic polynomial.