Let $\pi\colon{P}\to{M}$ be a principal $G$-bundle and $\Omega_{hor}^{k}\left(P,\mathfrak{g}\right)$ the space of $\mathfrak{g}$-valued horizontal${}^*$ differential forms. Given a global section $s\in\Gamma\left(P\right)$, does the pullback of $s$ define an isomorphism between $\Omega_{hor}^{k}\left(P,\mathfrak{g}\right)$ and $\Omega^{k}\left(M,\mathfrak{g}\right)$ ? That is, is the map: $$\Omega_{hor}^{k}\left(P,\mathfrak{g}\right)\ni\omega\mapsto{s^*}\omega\in\Omega^{k}\left(M,\mathfrak{g}\right)$$ an (vector bundle) isomorphism?
Intuitively, I expect the answer to be "yes". If this is the case, can you please provide a sketch of a proof?
$*$ Horizontal forms are, by definitionn, those that are identically zero when at least one of their arguments belongs in $T^{v}P=:\ker{\pi_*}$
This isn't true. It's possible to have $s^* \omega = 0$ without $\omega$ being zero (say, start with any non-zero horizontal form and then multiply it by a bump function which is zero around the image of the section $s$).
The correct statement goes as follows. Consider $\Omega^k_\text{bas}(P, \mathfrak{g})$ to be the basic forms, i.e. horizontal forms which satisfy $R_g^* \alpha = \text{Ad}_{g^{-1}} \alpha$ for all $g \in G$, where $R_g : P \to P$ is translation by $g$. You can show that the pullback via $\pi : P \to M$ of a form in $\Omega^k(M, \mathfrak{g})$ in fact gives you a basic form, so that you have a map
$$ \pi^{*} : \Omega^k(M, \mathfrak{g}) \to \Omega^k_\text{bas}(P, \mathfrak{g})$$
and this map is an isomorphism. The proof is pretty straightforward, but for a sketch see proposition 2.49 in page 26 of these notes.