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Two ways of thinking about $L_X$ on differential forms:

(1) Define it by using the infinitesimal flow determined by $X$. This implies that (a) $L_X$ is a degree $0$ derivation of the algebra of differential forms (because pulling back by a diffeomorphism is an automorphism of the algebra), and (b) it commutes with $d$ (because pulling back by a diffeomorphism commutes with $d$), and (c) $L_Xf=Xf$. And there cannot be more than one operator with properties (a), (b), (c), because the algebra is generated by functions and closed exact $1$-forms.

(2) Define it by $L_X=d\circ i_X+i_X\circ d$. This also implies (a), (b), and (c) (using $d^2=0$), so it has to be the same as (1).

Conceptually it's better to think of an operator on sheaves of differential forms (forms defined on open subsets), because the fact the algebra of global forms is generated by what we said uses some ad hoc constructions in the $C^{\infty}$ case and is false in the holomorphic case, or in algebraic geometry. And of course in algebraic geometry you don't have the flow even locally, so (2) is especially good.

I did not mention tangent vector fields and $L_XY=[X,Y]$. But let me point out the following.

For any sort of tensor bundle you can name (made by starting with tangent bundle, dualizing, tensoring, symmetrizing, ), there is an $L_X$ acting. These are all instances of the following: you have a derivation $D$ of functions, and a related linear operator $L$ on sections of the bundle, and it satisfies $L(fs)=(Df)s+fLs$. The operator on the tensor product of two bundles satisfies $L(s\otimes t)=Ls\otimes t+s\otimes Lt$. And the operator on a bundle and its dual? If we denote the pairing of sections of $E$ and sections of $E^\star$ into functions by $(s,\alpha)$, then we have $D(s,\alpha)=(Ls,\alpha)+(s,L\alpha)$. This can serve as definition of either $Ls$ or $L\alpha$ in terms of the other.

For $E=TM$ and $D=X$ and $L=L_X$ this says $X(Y,df)=(L_XY,df)+(Y,L_Xdf)$, i.e. $XYf=(L_XY)f+(Y,L_Xdf)$, which makes the statement $L_XY=XY-YX$ equivalent to the statement $L_Xdf=d(Xf)$.

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Two ways of thinking about $L_X$ on differential forms:

(1) Define it by using the infinitesimal flow determined by $X$. This implies that (a) $L_X$ is a degree $0$ derivation of the algebra of differential forms (because pulling back by a diffeomorphism is an automorphism of the algebra), and (b) it commutes with $d$ (because pulling back by a diffeomorphism commutes with $d$), and (c) $L_Xf=Xf$. And there cannot be more than one operator with properties (a), (b), (c), because the algebra is generated by functions and closed $1$-forms.

(2) Define it by $L_X=d\circ i_X+i_X\circ d$. This also implies (a), (b), and (c) (using $d^2=0$), so it has to be the same as (1).

Conceptually it's better to think of an operator on sheaves of differential forms (forms defined on open subsets), because the fact the algebra of global forms is generated by what we said uses some ad hoc constructions in the $C^{\infty}$ case and is false in the holomorphic case, or in algebraic geometry. And of course in algebraic geometry you don't have the flow even locally, so (2) is especially good.

I did not mention tangent vector fields and $L_XY=[X,Y]$. But let me point out the following.

For any sort of tensor bundle you can name (made by starting with tangent bundle, dualizing, tensoring, symmetrizing, ), there is an $L_X$ acting. These are all instances of the following: you have a derivation $D$ of functions, and a related linear operator $L$ on sections of the bundle, and it satisfies $L(fs)=(Df)s+fLs$. The operator on the tensor product of two bundles satisfies $L(s\otimes t)=Ls\otimes t+s\otimes Lt$. And the operator on a bundle and its dual? If we denote the pairing of sections of $E$ and sections of $E^\star$ into functions by $(s,\alpha)$, then we have $D(s,\alpha)=(Ls,\alpha)+(s,L\alpha)$. This can serve as definition of either $Ls$ or $L\alpha$ in terms of the other.

For $E=TM$ and $D=X$ and $L=L_X$ this says $X(Y,df)=(L_XY,df)+(Y,L_Xdf)$, i.e. $XYf=(L_XY)f+(Y,L_Xdf)$, which makes the statement $L_XY=XY-YX$ equivalent to the statement $L_Xdf=d(Xf)$.