I'm trying to understand the Lie bracket operation $[X, Y]$ as the rate of change of $Y$ as seen by an observer moving along the flow of $X$.
Example 1
Suppose $X=\{1, x\}^T$ and $Y=\{1, 0\}^T$, then $[X, Y]=\{0, -1\}$ and the flow of $X$ is $\{t+x_0,\frac{1}{2}t^2+ t\ x_0+ y_0\}$.
But I'm failing to see how along any of the flows below, the rate of change of vector $\{1, 0\}$ is $\{0,-1\}$.
Example 2
Suppose $X=\{x, x\}^T$ and $Y=\{1, 0\}^T$, then $[X, Y]=\{-1, -1\}$ and the flow of $X$ is $\{e^t\ x_0,e^t\ x_0-x_0+y_0\}$.
Again from plot below I'm not able to see how the directional derivative makes sense.
Edit
Adding except from the source.
Too long for a comment.
$Y=\{1,0\}^\top$ which I prefer to write as $Y=\partial_y$ is a constant vector field. Its directional derivative w.r.t. any other vector field $X$ is zero. The Lie dervative ${\cal L}_XY$ of a vector field $Y$ w.r.t. $X$ is the difference of the directional derivatives: $$ {\cal L}_XY=[X,Y]=\partial_XY-\partial_YX\,. $$
In your first case, \begin{align} X&=\partial_x+x\partial_y\,,&Y&=\partial_x\,,&XY&=\partial_x^2+x\partial_y\partial_x\,,&YX&=\partial_x^2+\partial_y+x\partial_x\partial_y \end{align} so that the commutator becomes \begin{align}\require{cancel} [X,Y]&=XY-YX=\cancel{\partial_x^2}+\bcancel{x\partial_y\partial_x}-\cancel{\partial_x^2}-\partial_y-\bcancel{x\partial_x\partial_y}=-\partial_y\,. \end{align} On the other hand, the directional derivatives are calculated by applying $\partial_x,\partial_y$ to the components of $X$ and $Y$ and taking the dot product with the components of $Y$, resp. $X\,:$ $$ \partial_XY^\nu=X^x\partial_xY^\nu+X^y\partial_yY^\nu\,. $$ This is the $\nu$-th component of $\partial_XY\,.$
In your first case $X^x=1,X^y=x,Y^x=1,Y^y=0$ so that \begin{align} \partial_XY&=0\,,&\partial_YX=\partial_y\,. \end{align}