Let $\phi : G \times M \rightarrow M$ be a group action on a smooth manifold $M$ and Lie group $G$.
Then we define $$f(t):=\phi(g(t),d(t)).$$
where $g: I \rightarrow G$ and $d: I \rightarrow M$ are smooth maps.
Now I'd say: $$f'(t) = D\phi(g(t),d(t))(g'(t),d'(t)).$$
In the lecture we wrote: $$f'(t) = D \phi_{g(t)} d(t) d'(t) + D \phi_{g(t)} (d(t)) X_{\zeta(t)} (d(t))$$ where $\zeta(t) = dL_{g(t)^{-1}}g'(t)$ is an element of the Lie Algebra, $dL$ is the derivative of the left-translation and $X_{\zeta}(p):=\frac{d}{dt}|_{t=0} \phi_{e^{t \zeta}}(p)$ for any $\zeta$ in the Lie Algebra. The thing is that I don't have the slightest clue, how one could come up with this second expression. Could anybody explain to me whether I am wrong or give me an idea why both expressions might be equivalent?
So if you had no idea how it worked, why didn't you ask in the lecture?
First, split the problem up (I will use the letter $u$ instead of $d$ because of possible confusion with the differential) into the question how to differentiate the function $f(t) = \phi(g, u(t)) = g \cdot u(t)$ for fixed $g \in G$, and how to differentiate the function $f(t) = \phi(g(t), p) = g(t) \cdot p)$ for a fixed point $p \in M$.
The first is really straightforward. $g$ acts as a diffeomorphism $\phi_g$ of $M$, so we just have to calculate $$ f^\prime(t) = \frac{\partial}{\partial t} \bigl\{ \phi_g(u(t))\bigr\} = D \phi_g\big|_{u(t)} u^\prime(t)$$ Here $D\phi_g$ is just the differential of the diffeomorphism $\phi_g$, and it has to be evaluated at the point $u(t)$.
Now the second case can be reduced to the first case. Namely, suppose you want to differentiate at the point $t_0$ and write $h_t := g_{t_0}^{-1} g_t$ and $u(t) = g_{t_0}^{-1} g_t \cdot p$. Then clearly $g_t \cdot p = g_{t_0} \cdot u(t)$, so that $$f^\prime(t_0) = D\phi_{g_{t_0}}\big|_{u(t_0)} u^\prime(t_0) = D\phi_{g_{t_0}}\big|_{p} u^\prime(t_0).$$ It remains to calculate $u^\prime(t)$. We have $$h^\prime(t_0) = \frac{\partial}{\partial t}\Big|_{t=t_0} g_{t_0}^{-1} g_t = dL_{g_{t_0}^{-1}} g^\prime(t) = \xi(t_0)$$ where $\xi(t_0)$ is an element of the Lie algebra $\mathfrak{g} = T_eG$. Hence $h_t = g_{t_0}^{-1}g_t$ is a curve starting at $e \in G$ at time $t_0$ with velocity $\xi$. $\exp(t\xi)$ is another such curve. Hence $$u^\prime(t_0) = \frac{\partial}{\partial t}\Big|_{t=t_0}g_{t_0}^{-1}g_t\cdot p = \frac{\partial}{\partial t}\Big|_{t=t_0} e^{t\xi}\cdot p = X_\xi(p)$$ The claim follows.