Let $\omega$ be a vector-valued 1-form where the vector space is the Lie algebra $\mathfrak g$ and let $\mathcal P$ be a trivial principal bundle over a manifold $M$. Thus
$$ \omega\in\Omega^1(\mathcal P,\mathfrak g)=\Gamma((\mathcal P \times \mathfrak g)\otimes T^*\mathcal P).\tag1 $$
I am reading that pulling back $\omega$ via the trivializing section $\sigma:M\to\mathcal P$ gives me a pull-back section $\sigma^*\omega\in\Omega^1(M,\mathfrak g)=\Gamma((M\times\mathfrak g)\otimes T^*M)$. Namely, $\sigma^*\omega$ is a smooth section of the bundle $(M\times\mathfrak g)\otimes T^*M$.
Since a pull-back section is a section in the pull-back bundle, namely
$$ \sigma^*\omega\in\Gamma(\sigma^*((\mathcal P\times\mathfrak g)\otimes T^*\mathcal P)),\tag2 $$
if what I am reading is correct, it follows that
$$ \sigma^*((\mathcal P\times\mathfrak g)\otimes T^*\mathcal P)\cong(M\times\mathfrak g)\otimes T^*M. $$
Moreover, given a fiber bundle $\pi:E\to M$ and a function $f:M'\to M$, then the pull-back bundle $f^*E$ and the bundle $E$ share same fibers over points $x'\in M'$ and $x=f(x')\in M$, since it is a special case of a product bundle.
Then it follows that $(\mathcal P\times\mathfrak g)\otimes T^*\mathcal P$ must share same fibers with $(M\times\mathfrak g)\otimes T^*M$
$\textbf {Problem}$: same fibers means $\mathfrak g\otimes T^*P\cong\mathfrak g\otimes T^*M$, which is of course not true. So what's wrong?