This problem is exactly same problem in Chern-Weil Theorem. Here I focus simplest examples, say trace without considering the shape of $P_m(F_1, F_2, \cdots, F_m)$.
Let $A_t = A_0 + t\eta$, $F_0 = dA_0 + A_0 \wedge A_0$, $F_t = dA_t + A_t \wedge A_t $ Decompose $F_t$ into $F_0$ I have \begin{align} F_t = F_0 + t(d\eta + A_0 \wedge \eta + \eta \wedge A_0) + t^2 \eta \wedge \eta \end{align}
Now consider $ tr(F_t^m)$.
What I want to prove is following
\begin{align} \frac{d}{dt} tr(F_t^m) = m \phantom{1} d [ tr(\eta \wedge F_t^{m-1})] \end{align} Let's call this equation (*)
Followings are my trial \begin{align} \frac{d}{dt} tr(F_t^m) &= m \phantom{1} tr\left(\frac{\partial F_t}{\partial t} F_t^{m-1}\right) \\ & = m \phantom{1} tr\left( (d\eta + A_0\wedge \eta + \eta \wedge A_0) \wedge F_t^{m-1}\right) + 2 m t \phantom{1} tr\left(\eta \wedge \eta \wedge F_t^{m-1}\right) \end{align}
Now \begin{align} md \phantom{1} tr(\eta \wedge F_t^{m-1}) = mtr(d\eta \wedge F_t^{m-1}) - m(m-1) tr(\eta \wedge d F_t \wedge F_t^{m-2}) \end{align} Using bianchi identity $dF_t + [A_t, F_t]=0$, I can replace the last term into $-[A_t, F_t]$, but still I have no idea that R.H.S and L.H.S are same in equation (*)
Note that this $A, F$ are Lie algebra valued differential forms. Note for $p$, $q$ differential forms we have \begin{align} w_p \wedge w_q = (-1)^{pq} w_q \wedge w_p \end{align} But for Lie algebra valued $p$, $q$ forms \begin{align} A_p \wedge A_q = (-1)^{pq +1} A_q \wedge A_p \end{align}
using this I have \begin{align} &tr(\eta \wedge \eta \wedge F_t^{m-1}) = - tr(\eta \wedge F_t^{m-1} \wedge \eta)=0 \\ & tr(\eta \wedge A_0 \wedge F_t^{m-1}) = tr(A_0 \wedge \eta \wedge F_t^{m-1}) = - tr(A_0 \wedge F_t^{m-1} \wedge \eta)=0 \end{align} and this guarantee the equation (*).