Given a G right action on a principal bundle P, $\triangleleft: P \times G \rightarrow P$, if we have a curve $\gamma(t) = \delta(t) \triangleleft g(t)$ for $\gamma ,\delta \in P, g \in G$, I would guess the corresponding tangent vector to $\gamma$ is:
$$ \gamma'(t) = (\delta \triangleleft g)'(t) = (\triangleleft g)_*\delta'(t) $$
Based on the definition of the pushforward, and where the slight abuse of notation $\triangleleft g : P \rightarrow P$ indicates the right action, and $\triangleleft g_*$ is its corresponding pushforward.
However, in Theorem 5.8.2 of Hamilton's "Mathematical Gauge Theory", it is stated that (in the context of horizontal lifts) the tangent vector to $\gamma$ is:
$$ \gamma'(t) = (\triangleleft g)_*\delta'(t) + \tilde\Xi(g'(t)) $$
Where $\Xi$ is the Maurer-Cartan form and the tilde denotes the fundamental vector field, ie for a Lie algebra element X, the corresponding fundamental vector field is $\tilde X_p = (p\triangleleft \exp(tX))'(t)$
I'm not sure why this second term is necessary, and why my original equation is wrong. The theorem in "Mathematical Gauge Theory" refers to another theorem 3.5.4, which seems to manipulate the pushforward of the map $\triangleleft: P \times G \rightarrow P$, but once again I'm not sure how this works and why it is necessary.
It seems what I was missing was that g was also a curve, and therefore the Leibniz formula should have been applied. Following proposition 1.4 of Kobayashi & Nomizu's "Foundations of Differential Geometry Vol 1", If we have two manifolds $X,Y$ with $x(t)\in X, y(t)\in Y$, then the cartesian product of these two manifolds will have $z(t):= (x(t), y(t))\in X\times Y$. My mistake was assuming the corresponding tangent vector at $t = t_0$ was $z'(t_0) = (x'(t_0), y'(t_0))$, however actually apply $z(t_0)$ on an arbitrary function $f:X\times Y \rightarrow \mathbb{R}$, then
$$\begin{aligned} z'(t_0)f &= \frac{d}{dt}\bigg\rvert_{t=t_0}f(z(t)) = \frac{d}{dt}\bigg\rvert_{t=t_0}f(x(t), y(t))\\ &=\lim_{\Delta t \rightarrow 0} \frac{f(x(t_0+\Delta t),y(t_0+\Delta t)) - f(x(t_0),y(t_0))}{\Delta t}\\ &=\lim_{\Delta t \rightarrow 0} \frac{f(x(t_0+\Delta t),y(t_0+\Delta t)) - f(x(t_0),y(t_0+\Delta t)) + f(x(t_0),y(t_0+\Delta t)) - f(x(t_0),y(t_0))}{\Delta t}\\ &=\frac{d}{dt}\bigg\rvert_{t=t_0}f(x(t), y(t_0)) + \frac{d}{dt}\bigg\rvert_{t=t_0}f(x(t_0), y(t))\\ &= \bar{x'(t_0)}f + \bar{y'(t_0)}f \end{aligned}$$
where $\bar{x'(t_0)} = (x(t),y(t_0))'$ and similarly for $\bar{y'(t)}$. The second term involving the Maurer Cartan form therefore comes from this rule.