Where does this order parameter come from? I am having trouble understanding how one gets the order parameter term in the following.

The author says that we have to use Taylor series. Doing so, I get
$$
h \cdot \big(\nabla (f(x+h)-f(x))\big) \,=\, h \cdot \big(\nabla (h \cdot \nabla f (x)+ o(h^2)\big).
$$
I think that from here the author applied grad throughout using $\nabla o(h^2) = o(h^2)$. However, I do not see why this should be true.
Could someone clarify how one gets the desired Taylor expansion?
 A: Your mistake is that you are interpreting $\nabla f(x+h)$ wrong. This is the gradient of $f$ evaluated at the point $x+h$ so you can't expand $f(x+h)$ and then apply $\nabla$ (which is what you did), rather you must taylor expand the whole thing as a multivariable vector valued function. Perhaps doing it in one dimension first would help. But I will do the multivariable directly here.
The multivariable expansion of a function $g$ (where $g$ is vector valued) is:
$$g(x+h)=g(x)+\nabla g(x) \cdot h +O(h^2)$$
Now apply it to the function $g(x)=\nabla f(x)$
$$\nabla f(x+h)=\nabla f(x) + \nabla(\nabla f)(x)\cdot h +O(h^2) $$
A: Just literally the normal Taylor and nothing else. Also note that it is "big Oh" rather than "little oh", i.e. $O(h^2)$ and not $o(h^2)$".
Try to ignore the $\nabla$ and write say $g = \nabla_i f$. Now you have
$$
g(x+h) = g(x) + h\cdot \nabla g(x) + O(h^2)
$$
Then replace $g$ with $\nabla_i f$:
$$
\nabla_i f(x+h) = \nabla_i f(x) + h\cdot \nabla \nabla_i f + O(h^2).
$$
Rearrange and multiply by another $h$ to get the first expression.
