Hazard Function The rate is the expected number of events per some unit (time, or spatial). $\text{Rate}=\frac{\text{numbers of events}}{\text{the length of time interval}}.\ldots (1)$ 
(Reference: https://stats.stackexchange.com/users/11887/kjetil-b-halvorsen).
The definition of hazard function (or, hazard rate) is "the instantaneous rate of death at time $t$, given that the individual survives up to $t$".
The hazard function $h(t)$ can be defined as
$$h(t)=\lim_{\Delta t\to 0} \frac{Pr(t\le T<t+\Delta t|T\ge t)}{\Delta t}.\ldots (2)$$
(Reference: Lawless, J. F. (2011). Statistical models and methods for lifetime data (Vol. 362). John Wiley & Sons.).
My question is why $h(t)$ is not defined as $\frac{\text{NUMBER of death at time t, given that the individuals survive up to t}}{\Delta t}$?
More Specifically, why is the numerator of equation (2) probability? Why is it not number of death?
 A: You are confusing two very different things. The definition of $h(t)$, which is just a deterministic function $h:[0,+\infty]\to[0,+\infty]$, and the estimator $\hat{h}(t)$ of $h(t)$, which is a statistic, i.e., a function of your random sample from the population, and thus a random variable itself1.
The definition of hazard rate is what you report (not sure why the dots, I just removed them):
$$h(t)=\lim_{\Delta t\to 0} \frac{Pr(t\le T<t+\Delta t|T\ge t)}{\Delta t}$$
Now, suppose that for a given problem you don't know the expression of $Pr(t\le T<t+\Delta t|T\ge t)$, so you can't compute the above limit. If you can sample from the population, you can estimate it. Now, we need to show that 
$$\hat{h}(t) = \frac{\text{number of deaths at time } t}{\text{number of individuals at risk at time } t}$$
is a reasonable estimator of $h(t)$. 
Basically, the problem here is that $\text{number of individuals at risk at time } t$ is a really confusing terminology and the authors shouldn't have used it. Anyway, since, given a fixed time $t$, no individual dies exactly at $t$, the numerator must be intended as $\text{number of deaths between time } t \text{ and time } t+\delta t$, where $\delta t$ is suitably small (it's an estimate, after all). Of course, only people who were still alive at time $t$ could die in $[t, t + \delta t]$. Now, the ratio   
$$\hat{Pr}(t\le T<t+\delta t|T\ge t) = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{\text{number of individuals alive at time } t}$$
is a frequentist estimate of $Pr(t\le T<t+\delta t|T\ge t)$: this seems intuitive to me, but I add a detailed explanation as requested by the OP. Feel free to skip this part.

Let $T$ be the random variable expressing the time of death of an individual. We want to estimate $Pr(t\le T<t+\delta t)$. We could then draw a random sample $D=\{T_1,\ldots,T_N\}$ of iid random variables, which in practice means following a group of similar individuals until they die. Of course, since you can only die once, we cannot follow the same individual across multiple lives, as opposed to when we throw the same coin multiple times to estimate $Pr(\text{Heads})$. We need to follow a group of individuals, instead. Now, as explained in introductory probability courses, an estimate of $Pr(t\le T<t+\delta t)$ based on $D$ is just
$$\hat{Pr}(t\le T<t+\delta t) = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{N}$$
However, we really want to estimate 
$$Pr(t\le T<t+\delta t|T\ge t) = \frac{Pr((t\le T<t+\delta t) \cap (T\ge t))}{Pr(T\ge t)} $$
The estimate for the denominator is obvious: someone will die at a time $T\ge t$ iff he's alive at time $t$, thus
$$\hat{Pr}(T \ge t) = \frac{\text{number of individuals alive at time } t}{N}$$
For the numerator, since the event $t\le T<t+\delta t$ is included in  the event $T \ge t$, $(t\le T<t+\delta t) \cap (T\ge t) = t\le T<t+\delta t$. Thus
$$\hat{Pr}(t\le T<t+\delta t|T\ge t) = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{\text{number of individuals alive at time } t} $$

Thus,  $\hat{h}(t)$ defined as 
$$\hat{h}(t)=\frac{\hat{Pr}(t\le T<t+\delta t|T\ge t)}{\delta t} = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{(\text{number of individuals alive at time } t) \times \delta t}$$
is a reasonable estimator of $h(t)$. I guess that the definition used by your book is trying to get at the same estimator, in a very convoluted way,. Probably, what the authors mean is that not all individuals alive at time $t$ are likely to die in a subsequent short time interval. Intuitively, the longer the time interval $\delta t$, the larger is the number of "at risk" individuals, among all individuals alive at time $t$. If we suppose that the number of individuals that, at time $t$, are at risk of death in $[t,t+\delta t]$, is equal to $(\text{number of individuals alive at time } t) \times \delta t$, then we get
$$\hat{h}(t) = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{\text{number of individuals at risk at time } t}$$
which is the definition reported by your book.

1 to be precise, $\hat{h}(t)$ is a random process, not a random variable: however, for each fixed $\bar{t}\in[0,+\infty]$, $\hat{h}(\bar{t})$ is a random variable.
A: By definition, the hazard rate is the rate of death of a population of arbitrary size. The hazard rate you propose - let us call it $\widetilde{h(t)}$ - would be defining what you are interested in implicitly as a function of the population size $S$. In particular, the relationship between the original $h(t)$ and your $\widetilde{h(t)}$ is given by $S \times h(t) =  \widetilde{h(t)}$. Naturally, you are more interested in a population-size independent function.
