Tensorality of the Lie Derivative and $i([X,Y]) = [L(X),i(Y)]$

I'm trying to understand the equation following (2.15) on p.9 of Blau's Symplectic Geometry and Geometric Quantization.

For two vector fields $$X,Y$$ on a symplectic manifold $$M$$ we are told one statement of the tensorality of the Lie derivative is

$$i([X,Y]) = L(X)i(Y) - i(Y)L(X)$$

where $$i(X)$$ denotes inclusion into the first argument of the tensor being acted upon and $$L(X)$$ denotes the Lie derivative of the tensor being acted upon along the vector field $$X$$.

Here's my current understanding: from the linearity of tensors we have

$$i([X,Y])= i(XY-YX) = i(XY) - i(YX).$$

I would expect the next step somehow writes e.g. $$i(XY)$$ as $$L(X)i(Y)$$ but I cannot quite see the justification for this, and anyway wouldn't the result then be

$$i([X,Y]) = L(X)i(Y) - L(Y)i(X)~?$$

Well, the composition $$XY$$ of the vector fields $$X$$ and $$Y$$ may be an operator on $$C^\infty(M)$$, it is however not a vector field! Indeed, vector fields are first order linear partial differential operators, like, for example, the partial derivatives $$\partial_x$$ and $$\partial_y$$ on $$C^\infty(\mathbf R^2)$$. Their composition $$\partial_x\partial_y$$ is not first order. Your equation $$i(XY-YX)=i(XY)-i(YX)$$ simply does not make sense since neither $$i(XY)$$ nor $$i(YX)$$ make sense.
The right way to derive the equation you want to understand is to consider a contraction $$i(Y)T$$ of a tensor $$T$$ as multplication $$Y\cdot T$$, and check that the Lie derivative of a contraction satisfies the Leibniz rule: $$L_X(Y\cdot T)=L_X(Y)\cdot T+Y\cdot L_X(T),$$ which translates as $$L_Xi(Y)=i([X,Y])+i(Y)L_X$$ since $$L_XY=[X,Y]$$. Voilà your formula!