Consider a birth process, where at each unit time step $t$, on the average, there is a fraction of $0.012$ among all the populations who will give a birth. Suppose the doubling time is $m$, then, from $1.012^m =2$ we get that $m$ is approximately $58.108$, i.e., the whole population will double after $58.108$ time steps. Then, the frequency, i.e., the average rate of events per time step is $1/58.108$. We can thus model it using a possion process with parameter $1/58.108$. Then, the probability for a birth-event within one time step is $1-e^{-1/58.108}$, which is approximately $0.0171$. This reasoning should be right, right?
Now, probability is a proportion in some sense. Note the fact that a proportion of $0.012$ would give a birth during each time step. So, is it safe to say that the probability of giving a birth during each time step is $0.012$ then? But that is not the same as $0.0171$, not even close. Can anyone tell me why? Any comments are greatly appreciated.
You need to be careful with your definition, because you've defined it as a discrete time process where individuals have either $1$ or $0$ births in each interval. But you're trying to model it as a continuous time process. Which is it?
Even if it is a continuous time process, it still doesn't make sense to use a time-homogeneous Poisson process (as you did, giving it a constant rate $1/58.108$). You will need to convince yourself that if individuals are each giving birth at a rate of $0.012$, then a larger population will have a larger population birth rate.
The thing which does this is called a pure birth process, or Yule process. This is different from a time-homogeneous Poisson process (constant rate $\lambda$) and different also from a time-inhomogeneous Poisson process (rate $\lambda(t)$ depends on time). In the pure birth process, the rate $\lambda_n$ depends on the number of individuals in the population. If you let the rate be equal to $\lambda \cdot n$ (i.e. a constant multiple of the population size) then the result is that the population size $N(t)$ is, unfortunately, not even a Poisson random variable. It turns out that $Pr[N(t) = n | N(0) = k]$ has a negative binomial distribution with parameters $k$ and $e^{\lambda t}$. Hope this helps!