In the literature, We define a Hardy field as being a set of germs of function defined on some positive half line of $\mathbb{R}$ which is a field closed under differentiation.
But in Miller's work (for example his article, Basics of o-minimality and hardy fields), he considers germs of functions defined on all of $\mathbb{R}$ .
If we adopt the first definition, Using the monotonicity theorem, I find it easy to show that If $\mathcal{R}$ is an o-minimal expansion of $\overline{\mathbb{R}}$ then the set of unary definable functions in $\mathcal{R}$ that are defined for all sufficiently large elements forms a Hardy field.
But if we adopt Miller's definition, Denote by $\mathcal{H}$ the set of germs of unary definable function on $\mathcal{R}$ , and let $[f]\in\mathcal{H}$ . to show that $\mathcal{H}$ is closed under differentiation, we need to find a definable function $g:\mathbb{R}\rightarrow\mathbb{R}$ such that $g$ is differentiable, $[f]=[g]$ and $[g']\in\mathcal{H}$.
Let $[f]\in\mathcal{H}$ . By the monotonicity theorem, there exists $a\in\mathbb{R}$ such that on $(a,\infty)$ $f$ is $C^1$ . Of course the derivative of a definable function is definable. But it is not defined on all of $\mathbb{R}$ . I know that there are theorems of extension. But I am not sure about the definability of such extensions!!!
You seemingly did not ask an explicit question in the text, but you write:
Mayhap I'm missing something, but I don't see much evidence in the lecture notes themselves that Miller really does want to work only with germs of everywhere defined functions specifically (see the comments below for some possible evidence, though!).
But it makes no difference. Given an o-minimal expansion of a real-closed field $R$ and a function $f:R\rightarrow R$, you can obtain an open interval $(a,+\infty)$ on which $f$ is differentiable, use density to pick a point $b$ in this interval, then you get a function $g: R \rightarrow R$ as
$$g(x) = \begin{cases} f'(b)x + f(b) - f'(b) b, & \text{if } x < b, \\ f(x), & \text{otherwise} \end{cases}$$
and now $g$ is definable, once-differentiable everywhere, while having the same germ at $+\infty$ as $f$.