Well, here's a more detailed account of how to solve that exercise:
You have $Y=-y\dfrac{\partial}{\partial x} +x\dfrac{\partial}{\partial y}$
To compute its flow $\phi^Y_t$ you need to solve the following set of coupled ODE's:
$$\dot{x}=-y$$ $$\dot{y}=x$$
This is not too hard to solve (just differentiate one of the equations and plug in the other one, you'll get a Harmonic oscillator-esque ODE). In any case the solution is (which you can explicitly check by plugging in):
$$x(t)=x_0cos(t)-y_0sin(t)$$
$$y(t)=x_0sin(t)+y_0cos(t)$$
Hence the flow is $$\phi^Y_t(x,y)=\big(xcos(t)-ysin(t),xsin(t)+ycos(t)\big)$$
I switched from $x_0,y_0$ to $x,y$ since the flow can take arbitrary initial conditions $x,y$ and trace out their evolution as a function of $t$.
Now compute $d\phi^Y_t$ in standard coordinates (i.e. in the identity chart); its matrix representation will be just the Jacobian matrix of partial derivatives hence:
$$ d\phi^Y_t=
\begin{pmatrix}
cos(t) & -sin(t) \\
sin(t) & cos(t) \\
\end{pmatrix}
$$
Hence $$(d\phi^Y_t)_{\phi^Y_{-t}(p)} \bigg(\epsilon \dfrac{\partial}{\partial x}\bigg)_{\phi^Y_{-t}(p)}=\begin{pmatrix}
cos(t) & -sin(t) \\
sin(t) & cos(t) \\
\end{pmatrix}\begin{pmatrix}
\epsilon \\
0 \\
\end{pmatrix}=\epsilon\begin{pmatrix}
cos(t) \\
sin(t) \\
\end{pmatrix}$$
The subscripts indicate "evaluated at" hence $(d\phi^Y_t)_{\phi^Y_{-t}(p)}$ means that it is $d\phi^Y_t$ evaluated at the point $\phi^Y_{-t}(p)$ (notice that it is constant as a function of $x,y$ hence we shouldn't worry about that but I included it for completeness).
$$(L_Y X)_p=\dfrac{d}{dt}\bigg|_{t=0} \bigg((d\phi^Y_t)_{\phi^Y_{-t}(p)} \bigg(\epsilon \dfrac{\partial}{\partial x}\bigg)_{\phi^Y_{-t}(p)}\bigg)=\begin{pmatrix}
0 \\
\epsilon \\
\end{pmatrix}$$
The reason I included $\epsilon$ should be clear from the post in your link; Also you can use other charts to compute $d\phi^Y_t$; for instance in the polar chart it would be the identity.
