I have a simple question on how to calculate $\phi_{t+s}=\phi_t \circ\phi_s=\phi_t(\phi_s)$.
For example let $\phi_t$ be defined as $$\phi_t(k_1,k_2)=(\frac{k_1}{1-k_1t},k_2 e^{-t}) $$ show that $\phi_{t+s}=\phi_t \circ\phi_s$.
I know that for the second component $\phi_t=e^{-t}$ so $\phi_t \circ\phi_s=e^{-t} e^{-s}=e^{-(t+s)}$. However in general I do not understand how to find the flow $\phi_t$ in problems like the above.
That is what is $\phi_t \circ\phi_s$ versus $\phi_t$.
$$\phi_t(k_1,k_2):=\left(\frac{k_1}{1-k_1t},k_2 e^{-t}\right) $$ Show that $\phi_{t+s}=\phi_t \circ\phi_s$.
\begin{align}\phi_t(\phi_s(k_1,k_2))&=\phi_t\big(\overbrace{\frac{k_1}{1-k_1s}}^{=:c_1},\overbrace{k_2 e^{-s}}^{=:c_2}\big)\\&=\left(\frac{c_1}{1-c_1t},c_2 e^{-t}\right)\\&=\left(\frac{\frac{k_1}{1-k_1s}}{1-t\frac{k_1}{1-k_1s}},k_2 e^{-s} e^{-t}\right) \\ &=\left(\frac{k_1}{(1-k_1s)(1-t\frac{k_1}{1-k_1s})},k_2 e^{-s} e^{-t}\right)\\ &=\left(\frac{k_1}{1-k_1s-k_1t},k_2 e^{-s} e^{-t}\right)\\&=\left(\frac{k_1}{1-k_1(t+s)},k_2 e^{-(t+s)}\right)\\&=\phi_{t+s}(k_1,k_2)\end{align}
Note that your example can be split in to two independent flows so you need only consider each independently. Now using the definition of function composition you are trying to show that starting from some initial points $x_0,y_0$ you have that $\phi_t (\phi_s (x_0,y_0 ))= \phi_{t+s}(x_0,y_0)$, so lets try it, we have \begin{align*} \phi_s (x_0,y_0) = \left( \frac{x_0}{1-x_0s},y_0 e^{-s}\right) \end{align*} Now lets apply it again to that new point, i.e. we consider \begin{align*} \phi_t\left( \frac{x_0}{1-x_0 s}, y_0 e^{-s}\right) = \left( \frac{\frac{x_0}{1-x_0s} }{{1 - \frac{x_0}{1-x_0s}t }} , y_0 e^{-s} e^{-t}\right) \end{align*} and I then claim that this is equivalent to \begin{align*} \left( \frac{x_0} {1 - x_0(s + t) }, y_0 e^{-(s+t)}\right) = \phi_{t+s}(x_0,y_0). \end{align*} I will let you check these results, note there are other properties you must check to show that this is a flow, but these are easy!