What does ad$f$ mean, for $f$ a smooth function? I am currently reading Nicole Berline "Heat Kernels and Dirac Operators". On page 64 Differential Operators are introduced that are generalized from operators acting on scalar functions to vector bundles. 
Thereby it is mentioned that if $D$ is an i-th order operator and $f$ is a smooth function, then (ad $f)^i D$ is a zeroth order operator.
I am not sure I understand this comment, in case anybody can help that would be great!! Please let me know in case more details are necessary.

A function determines a 0th order differential operator by pointwise multiplication.  Then $(\text{ad } f) (D) = [f,D]$ is the commutator of the operator corresponding to $f$ with the operator $D$.
The statement that $(\text{ad } f)^i D$ is 0th order if $D$ is of order $i$ follows from the Leibniz (product) rule.
