His rule is better (stronger) because it tells you that the derivative commutes with taking "contractions" between tensors (e.g. $(\omega,Y)\mapsto \omega(Y)$)
You didn't specify what type of tensors your $T$ and $S$ are, but if they are say covectors, then $T\otimes S$ is a $(0,2)$-tensor (or maybe $(2,0)$, I don't know his convention), which takes in two vectors as input.
Now say $v,w$ are two vectors at a point, and choose extensions to vector fields which satisfy $\nabla v = 0,\nabla w = 0$ (you can see that derivatives on $v,w$ will cancel out if you carry them through, so the choice of extension doesn't matter).
Then, since $(T\otimes S)(v,w) = T(v)\cdot S(w)$, his rule says
\big(\nabla (T\otimes S)\big)(v,w)
&= \nabla\big((T\otimes S)(v,w)\big) - (T\otimes S)(\nabla v,w) - (T\otimes S)(v,\nabla w)\\
&= \nabla\big(T(v)\cdot S(w)\big)\\
&= \big[\nabla\big(T(v)\big)\big]\cdot S(w) ~+~ T(v)\cdot \big[\nabla\big(S(w)\big)\big] \\
&= \big((\nabla T)(v)\big)\cdot S(w) ~+~ T(v)\cdot \big((\nabla S)(w)\big) \\
&= \big((\nabla T)\otimes S\big)(v,w) ~+~\big(T\otimes (\nabla S)\big)(v,w)
so it implies your Liebniz rule in this case.
Note that he doesn't say that this "is equivalent" to what you say is the Liebniz rule, he just says that his axiom "is called the Liebniz rule".