The lecturer here wants the viewer to derive the components of the Lie derivative of a (1,1) tensor-field. But even before that I have a little question about the components of the Lie derivative of a vector field:
Careful: 1) and 2) are incorrect!
let $(U,x)$ be a chart and $X,Y$ vector fields on the smooth manifold $(M,\mathcal{O},\mathcal{A})$, I get:
$$ ({L}_X Y)^i = [X,Y]^i = (XY - YX)^i = X^m \left(\frac{\partial}{\partial x^m}\right) Y^i - Y^m \left(\frac{\partial}{\partial x^m}\right) X^i $$ 1) is that correct? I suspect, since we write single components, which are real functions, that I can reorder the terms, as I please (commutativity of multiplication on $C^{\infty}M$)? And the derivatives, which are actually the basis vectors, act on the function to which this thing is applied to anyway, right? So for clarity I could move them to the far right to show this: $$ = X^m Y^i \left(\frac{\partial}{\partial x^m}\right) - Y^m X^i \left(\frac{\partial}{\partial x^m}\right) $$ 2) still correct?
3) Then is there any "rule"/intuition or something why the contraction with the basis is over the "outer" field? If I take $(y \circ x^{-1})^i$ I can write this as $(y^i \circ x^{-1})$, basically the last function is responsible for the component I get out. In the above example it seems to be the first one applied.
Note that the components $X^i, Y^i$ of a vector field wrt some coordinates are functions. With $[X,Y]^i$ the i-th component of the Lie derivative is meant, this is not a vector but a function. As such in part 1) the derivatives act only on $X^i$ and $Y^i$, and not on the function in the argument of $[X,Y](f)$. Writing it out:
$$[X,Y]^i=X^m \frac{\partial Y^i}{\partial x^m}- Y^m \frac{\partial X^i}{\partial x^m}$$
or
$$[X,Y](f)=X(Y(f))-Y(X(f))= \left(X^m \frac{\partial Y^i}{\partial x^m}- Y^m \frac{\partial X^i}{\partial x^m}\right)\frac{\partial f}{\partial x^i} $$