Let $f: X \to X$ be a monotone function/operator with $X$ a Hilbert space.
Then $f$ is classically monotone or monotonically increasing if:
$$x \leq y \implies f(x) \leq f(y)$$ where $\leq$ represent an elementwise-$\leq$
Then $f$ is monotone if: $$\langle x - y, f(x) - f(y) \rangle \geq 0$$
Is there any connection (i.e. $\iff$) between the two concepts?
We can't necessarily say that $y\ge x$ any more.... although we could certainly say that the norm of one is greater than the other. In any case, if we restrict ourselves to the case when our operator is an ordinary linear map $f:\mathbb{R}\to \mathbb{R}$ then it certainly does make sense to say that if $y\ge x$ then $y-x\ge 0$ and we would require that $f(x)\le f(y)$ for the ordinary function to be monotonically increasing. From here we generalize, because $f(x)\le f(y)$ it certainly follows that $f(y)-f(x)\ge 0$. If we take the usual product then we will get
$$(y-x)\cdot (f(y)-f(x)) \ge 0$$
With normal functions and normal numbers this is an equivalent definition of a monotonically increasing function (indeed if $y-x\le 0$ then $f(y)-f(x)$ should be $\le 0$ if it is an increasing function).
Now to generalize this definition when $x\le y$ may not make sense so to fix this we use the definition that you gave, and it captures the special case (the classic definition).