Time derivative of a pullback of a time-dependent 2-form

This is a frequently-used trick in differential geometry. The key idea is the following fact, taken from basic multi-variable calculus:

Let $F:\mathbb{R}^2\to V$ be a smooth function, where $V$ is some vector space, and let $dF_p$ denote the differential of $F$ at $p$. Then at every $p\in\mathbb{R}^2$ the differential $dF_p$ is linear, and in particular, we have $$dF_p(1,1)=dF_p(1,0)+dF_p(0,1).$$

How is this related to the question at hand?

We wish to calculate the time derivative $\frac{d}{dt}\psi_t^*\omega_t$. The difficulty lies in the fact that $t$ appears twice in the expression we are trying to differentiate. So, the trick is to consider the two-parameter family $$F(t,s):=\psi_t^*\omega_s,$$ where $t$ and $s$ may vary independently. Then, by the above basic fact, we have $$\frac{d}{dt}\psi_t^*\omega_t=dF(1,1)=\frac{\partial F}{\partial t}+\frac{\partial F}{\partial s},$$ which gives the desired expression.

It is sort of Leibniz rule (as expressed from a more general and conceptual standpoint in Amitai Yuval's answer and as somewhat detailed below). Observe that $\psi^{\ast}_t\omega_t$ is a path in the vector space of differential forms. You could write

$$ \begin{array} \notag \left. \frac{d}{dt}(\psi^{\ast}_{t} \omega_t) \right|_{t = t_0} &= \lim_{t \to t_0} \frac{\psi^{\ast}_{t} \omega_t - \psi^{\ast}_{t_0} \omega_{t_0}}{t-t_0} = \lim_{t \to t_0} \frac{\psi^{\ast}_{t} \omega_t - \psi^{\ast}_{t_0} \omega_{t} + \psi^{\ast}_{t_0} \omega_{t} - \psi^{\ast}_{t_0} \omega_{t_0}}{t-t_0} \\ \notag &= \lim_{t \to t_0} \frac{\psi^{\ast}_{t} \omega_t - \psi^{\ast}_{t_0} \omega_{t}}{t-t_0} + \lim_{t \to t_0} \frac{ \psi^{\ast}_{t_0} \omega_{t} - \psi^{\ast}_{t_0} \omega_{t_0}}{t-t_0}\\ \notag &= \psi_{t_0}^{\ast} \left\lbrack \lim_{t \to t_0} \frac{(\psi_{t_0}^{\ast})^{-1}\psi^{\ast}_{t} - Id}{t-t_0}\, \omega_t \right\rbrack + \lim_{t \to t_0} \frac{ \psi^{\ast}_{t_0} \omega_{t} - \psi^{\ast}_{t_0} \omega_{t_0}}{t-t_0}\\ \notag &= \psi_{t_0}^{\ast} \left\lbrack \lim_{t \to t_0} \frac{(\psi_{t_0}^{\ast})^{-1}\psi^{\ast}_{t} - Id}{t-t_0} \right\rbrack \cdot \lim_{t \to t_0} \omega_t + \lim_{t \to t_0} \frac{ \psi^{\ast}_{t_0} \omega_{t} - \psi^{\ast}_{t_0} \omega_{t_0}}{t-t_0}\\ \notag &= \psi^{\ast}_{t_0} \mathcal{L}_{X_{t_0}}\omega_{t_0} + \psi_{t_0}^{\ast} \left. \frac{d}{dt} \omega_t \right|_{t_0}\, . \end{array} $$

In some regards, the real wonder is why we can write $\psi^{\ast}_{t_0} \mathcal{L}_{X_{t_0}}\omega_{t_0}$ in the last line. After all, Lie derivative and Cartan's formula are frequently introduced for flows, but the isotopy $\psi_t$ is not generally a flow as it does not satisfy $\psi_{t+s} = \psi_t \circ \psi_s$... Moreover, the appearance of $t$ both on $\psi$ and on $\omega$ in $\lim_{t \to t_0} \frac{\psi^{\ast}_{t} \omega_t - \psi^{\ast}_{t_0} \omega_{t}}{t-t_0}$ make its evaluation as done above a little more doubtful: one could feel uneasy with limits of operators on forms and as to why smoothness of the isotopy $\psi_t$ implies the smoothness (and thus the genuine differentiability) of the family of operators $\psi_t^{\ast}$.

To circumvent these problems, instead of thinking in an 'analytical way' as above, let's think geometrically. The isotopy of diffeomorphisms $\psi_t$ on the space $M$ induces a flow $\Psi_s$ on the space-time $M \times I$ (where $I$ is some open interval containing $[0,1]$) as follows: if $\psi_t$ was generated by integration of the $t$-dependent vector field $X_t$, then $\Psi_s$ is generated by integration of the $s$-independent vector field $Y(x,t) = X_t(x) + \partial/\partial t$ at $(x,t) \in M \times I$. Put differently, $\frac{d\Psi_s}{ds}(x,t) = X_t(x) + \partial/\partial t$. Similarly, the $t$-dependent form $\omega_t$ on $M$ is a $s$-independent form on $\Omega(x,t) = \omega_t(x)$ on $M \times I$.

Observe that $\Psi_t : M \times \{0\} \to M \times \{t\} : (x, 0) \to (\psi_t(x), t)$, hence the restriction of $\Psi_t^{\ast}\Omega$ to $M \times \{0\}$ is $\psi_t^{\ast}\omega_t$. Thus (dropping from the notation the point where everything is to be evaluated and using conventional abuse of notations, as it would get quite clumsy otherwise),

$$\begin{array} \notag \frac{d}{dt}(\psi^{\ast}_{t} \omega_t) &= \left. \frac{d}{dt}(\Psi^{\ast}_{t} \Omega \right|_{M \times \{0\}}) = \left. \Psi^{\ast}_{t}(Y \lrcorner \, d_{M \times I}\Omega + d_{M \times I}(Y \lrcorner \, \Omega) )\right|_{M \times \{0\}}\\ \notag &= \left. \Psi^{\ast}_{t}(Y \lrcorner \, (d_M\omega_{t} + d_I \Omega) + d_{M \times I}(X_t \lrcorner \, \omega_t))\right|_{M \times \{0\}} \\ \notag &= \left. \Psi^{\ast}_{t}(X_t \lrcorner \, d_M\omega_t + (\partial/\partial t) \lrcorner \, d_I \omega_t + d_{M}(X_t \lrcorner \, \omega_t))\right|_{M \times \{0\}} \\ \notag &= \left. \Psi^{\ast}_{t} (\mathcal{L}_{X_{t}}\omega_t + \frac{d \omega_t}{dt})\right|_{M \times \{0\}} \\ \notag &= \psi_t^{\ast}(\mathcal{L}_{X_{t}}\omega_t + \frac{d \omega_t}{dt}) \end{array}$$ For instance, as $\Omega = \sum_{ij} (\omega_t)_{ij} dx^i \wedge dx^j$, the notation $d_I \Omega$ means $\sum_{ij} \frac{d(\omega_t)_{ij}}{dt} dt \wedge dx^i \wedge dx^j$.