Let $\pi:P\rightarrow M$ and $\pi':P'\rightarrow M$ be principal $G$ and $H$ bundles respectively, and $f:G\rightarrow H$ be Lie group homomorphism. Let $F$ be a principal bundle homomorphism, that is a smooth map $F:P\rightarrow P'$ such that:
$$\pi'\circ F=\pi$$
and : $$F(p\cdot g)=F(p)\cdot f(g)$$ for all $p\in P$, and $g\in G$.
Suppose that $F$ is surjective, and $f$ is a surjective Lie group homomorphism. Furthermore, assume that $G$ and $H$ are connected for simplicity. Let $(U,\phi)$, be a principal bundle chart for $P'$, i.e. the trivialization $$\begin{align} \phi:\pi'^{-1}(U)&\longrightarrow U\times H\\ p&\longmapsto(\pi'(p),h(p)) \end{align}$$ where $h(p)$ is some smooth map $P'_U\rightarrow G$. Does it follow that there exists a principal bundle chart $(U,\psi)$ for $P$ such that:
$$\phi\circ F=(\text{Id}_U\times f)\circ \psi$$ If so, how would one construct $\psi$? If not, are there cases where this can happen? i.e. perhaps if $F$ and $f$ are double coverings? And if no such case exists, how can I see that this breaks down? Note that if $f$ is a surjective Lie group homomorphism, then $G/\ker f\cong H$, because the kernel is a closed Lie subgroup of $G$, and the action of $\ker f$ on $G$ is free and proper.
For context, I asked this question earlier today, and I believe if I can show that such a chart exists then I could easily prove what I seek out to prove. I am starting to suspect that no such chart exists however, I just can't see why. It seems that maybe such a chart would depend on global sections of $f$, which is a submersion.
In general, the setup you suggest tells you nothing about charts. Just consider the case $H=\{e\}$, then $\pi'=id_M:M\to M$ and $\pi:P\to M$ defines a bundle map for the trivial homomorphism $G\to\{e\}$, so you would conclude that $P$ has to be globally trivial.
The question can be analyzed as follows: As you noticed $\ker(f)$ is a normal subgroup of $G$. Hence you can form the orbit space $P/\ker(f)$ which is a smooth manifold and a principal bundle over $M$ with structure group $G/\ker(f)\cong H$. Moreover $P\to P/\ker(f)$ is a principal fiber bundle with structure group $\ker(f)$. Now $F$ induces an isomorphism of $H$-principal bundles from $P/\ker(f)\to P'$. Finally, you have to take into account that a chart of a principal bundle over $U\subset M$ is equivalent to a local trivialization of the bundle over $U$. So if you have a chart for $P'$ over $U$, this gives you a section $e:U\to P'$. Getting a corresponding chart for $P$ amounts to lifting this to a section of $P$, so the question is whether the restriction of the $\ker(f)$-principal bundle $P\to P/\ker(f)$ to $e(U)$ is trivial or not.
In the context of your original question with the spin groups, $\ker(f)=\mathbb Z_2$ and hence $P\to P/\ker(f)$ has to be a two-fold covering and hence trivial over $e(U)$ if $U$ is simply connected.