Let $G$ be a Lie group and $f = f(t, s) : \mathbb{R}^2 \to G$ smooth. Consider $\theta \in \Omega^1(G, \mathfrak{g})$ the Maurer-Cartan form of $G$. I'm trying to understand why
$\displaystyle\frac{d}{dt} \Bigr|_{t = 0} f^{*}\theta \left(\displaystyle\frac{d}{ds} \Bigr|_{s = 0}\right) = \displaystyle\frac{d}{ds} \Bigr|_{s = 0} f^{*}\theta \left(\displaystyle\frac{d}{dt} \Bigr|_{t = 0}\right) + \left[ \theta \left( \frac{df}{ds}(0, 0) \right), \theta \left( \frac{df}{dt}(0, 0) \right) \right]$
which is supposed to be a consequence of the Maurer-Cartan structure equation. The left hand side can be thought of as $\frac{d}{dt} \Bigr|_{t=0}$ acting on the differential of the function $f^{*}\theta \left( \frac{d}{ds} \Bigr|_{s = 0} \right)$, which is a function $\mathbb{R} \to \mathfrak{g}$. Here I suppose we could identify this differential as $f^{*}(d\theta)\left(\frac{d}{ds} \Bigr|_{s = 0}\right)$. Then the left hand side would be $f^{*}(d\theta)\left(\frac{d}{ds} \Bigr|_{s = 0}, \frac{d}{dt} \Bigr|_{t = 0}\right) = d\theta\left(\frac{df}{ds}(0, 0), \frac{df}{dt}(0, 0)\right) = -\left[ \theta \left( \frac{df}{ds}(0, 0) \right), \theta \left( \frac{df}{dt}(0, 0) \right) \right]$, but this result doesn't agree with the one stated above. I'm sure the error is in understanding the differential of $f^{*}\theta \left( \frac{d}{ds} \Bigr|_{s = 0} \right)$ but I've been confused on how to do it.