Related to this question for the background.
Let $\Omega$ be the curvature $2$-form if a principal bundle $E$ of frames (associated to a complex vector bundle $V$), let $\omega$ be the matrix valued connection on $E$. Assume that $\omega$ depends on a parameter $t$ (this is a passage for proving that the Chern Weil method doesn't depend on the connection used). By $\omega'$ I will denote the derivative with respect to the time. In the proof there is the following passage: $$ \text{tr}\{\Omega^n(\omega\omega'+\omega'\omega)\}= \text{tr}\{\Omega^n\omega\omega'-\omega \Omega^n\omega'\}$$ justified with the comment by the symmetry of trace.
Now it's unclear to me what symmetry we are referring to, I would simply notice that since $\omega$ is a $1$-form and $\Omega$ a $2$-form then $$\Omega^n\omega'\omega =- \omega\Omega^n\omega'$$ since first we swap order of $\omega,\omega'$ introducing a $-1$ in front and then swap order of $\omega,\Omega^n$ which doesn't not introduce any sign change being $\Omega$ a $2$-form.
Is that correct? what is this symmetry mentioned by the book?
You are correct about the negative sign, but wrong about the equality: note that both $\omega$ and $\Omega$ are $\mathfrak{gl}_m(\mathbb C)$-valued differential forms. In terms of local coordinates,
$$\omega = \omega_i^j, \ \ \ \Omega = \Omega_i^j.$$
Now write $A = \Omega^n \omega'$, where $A = A_i^j$. Then
$$(\Omega^n \omega' \omega)_i^j = (A\omega)_i^j = A_i^k \omega_k^j,\ \ \ (\omega \Omega^n \omega')_i^j = (\omega A)_i^j = \omega_i^k A_k^j = -A_k^j \omega_i^k.$$
(the -ve sign comes from the fact that both $A, \omega$ have odd degrees). So $\Omega^n \omega' \omega \neq -\omega\Omega^n \omega'$ unless $m=1$. But after taking trace,
$$\operatorname{tr}(\Omega^n \omega' \omega) = \operatorname{tr}(-\omega\Omega^n \omega') =\sum_{i,k} A_k^i \omega_i^k. $$
So by symmetry of trace, it is really the linear algebra fact that if both $A, B$ are square matrices, then $\operatorname{tr}(AB) = \operatorname{tr}(BA)$.