Differential in Lie groups I am trying to make sense of the Lie group machinery and relate it to the calculus. 
Suppose that $\psi(t)=\phi(s)\phi(t), s, t \in I$. 
Where $\phi(t)$ is a one-parameter subgroup of the Lie group $G$ and $I$ is an open interval containing $0$. 
So we have $\phi: \mathbb R \to G$ (a smooth homomorphism). 
Now using simple calculus we can find the differential of $\psi(t)$, treating it as a matrix, here each element is a function of $t$. Using the above, and for a fixed $s$, we can say as usual in the calculus $d\psi(t)=\phi(s)d\phi(t)$.
But if instead of calculus we use the Lie group definitions I cannot easily make sense of the differential. 
Based on the definition if $ \psi(t): \mathbb R \to G, t \in \mathbb R ,\psi(t) \in G$ (assuming the group structure of the manifold) then $d\psi(t)$ is
$d\psi(t): T_t\mathbb R \to T_{\psi(t)}G$ here, by definition we have 
$d\psi(t)(v)(g)=v(g \circ \psi) \in T_{\psi(t)}G, v\in T_t\mathbb R, g\in \mathcal F(G) $ 
Here $\mathcal F$ is the set of all smooth real-valued function on $G$. So $g: G \to \mathbb R$
Now my question is that how we can use the above definition of differential in Lie groups to say that 
$d\psi(t)=\phi(s)d\phi(t)$. 
or more explicitly,
$d\psi(t)(v)(g)=\phi(s)d\phi(v)(g)$
 A: I am going to write $\gamma := \phi(s)$, and $L_\gamma: G\to G$ for the map $x\mapsto\gamma x$. If I understand you correctly, you are assuming that $G$ is a matrix group, and your $\phi$ is a matrix with coefficients $\phi_{ij}$ which are functions of a real parameter $t$. Then it can be checked that $\dot\psi(t)\equiv(\gamma\phi)^\cdot(t) = \gamma\dot\phi(t)$, which in your notation is $d\psi(t) = \gamma d\phi(t)$. 
For the abstract definition, as @ACuriousMind remarked, you have to be careful with the notation. You are essentially asserting that $D_t(\gamma\phi) = \gamma D_t\phi$, but you have to be more careful. Our $\psi$ is the composition $L_\gamma\circ\phi$, and by the chain rule we have $D_t\psi = D_{\phi(t)}L_\gamma\circ D_t\phi$, so that what you really want to see is that in the case of matrix groups your $D_{\phi(t)}L_\gamma$ corresponds to left matrix multiplication by $\gamma$ on the Lie algebra, where its elements are represented by matrices of the same size. The most direct way to see this is essentially your same calculus computation.
