How confident can I be that a proportion I found in a sample is true of the population? Out of a population of roughly 2.0 million I took a random sample using the following parameters :
Confidence Level: 95%
Confidence Interval: 5
The sample I got (384 items as recommended by http://www.surveysystem.com/sscalc.htm ) did not have the characteristic I was looking for (lets call it purple iris)
So then I went and extracted a larger sample using this parameters:
Confidence Level: 95%
Confidence Interval: 1.58
That gives me a sample size of 3840. 
With this sample I was able to find 4 items with "purple iris". That is roughly .1%
So, my conclusion is that: I can be 95% confident that between (.1%-1.58) and (.1%+1.58) members of the population have "purple iris" which I believe means in plain English:

95 of every 100 populations like this will not have more than 2 of 100
  members with "purple iris"

Am I right? Is this the right way to estimate the proportion with which a characteristic appears in a population ?
 A: *

*Best-case scenario Overall, your sample size is 4224 (i.e., 384 + 3840) and your observed incidence is 4.  That gives you a two-sided 95%CI of around (0.026%, 0.24%) obtained via R's binom.test.  
Under this scenario, you could be 95% certain that the true population proportion fell within this interval.

*However, as @whuber comments, the sequence in which you took some sample, observed it, and then decided to take more sample may cause us to pause.  After your first sample, your CI on 0 successes out of 384 trials would have been (0.00%, 0.96%).  This is also a one-sided CI, so you could at that point have been 95% confident that the population incidence for "purple iris" was less than 1%.  
To my way of thinking, this result (throwing out the second, larger test) constitutes the worst-case scenario.  So your conclusion seems well supported (and could be stated it more strongly, using 1 instead of 2 out of 100, even within the worst-case scenario).
The "right" answer is probably somewhere between these two scenarios and depends on the rule for stopping sampling after the first round vs. moving to the second.  If you could formalize that, you might be able to estimate the probability of that rule getting invoked under different frequencies of the trait, and adjust your estimate accordingly.
A: A Bayesian approach would work well for this.
$$P(\theta|x,N)=\frac{P(x,N|\theta)P(\theta)}{P(x,N)}$$
In this case, $\theta$ is your proportion parameter, $x=4$ is the number of purple irises sampled, and $n=3456$ is the total number of samples (based on your comment that the sample size of 3840 included the first 348 samples). Your prior probability would be a Beta distribution: $P(\theta) = B(\theta|1,385)$; the parameters are the previously sampled number from each category + 1 (to codify the assumption that there is some chance of getting either type in a sample).  Using a Bernoulli likelihood,you can analytically calculate the posterior distribution of theta: $B(\theta|5,3837)$. 
This gives you a mean estimate of $\theta= 0.13$ and 95% credible intervals of $\{0.04\%, 0.27\%\}$.  Given that these are credible intervals and not confidence intervals, you could interpret this as 95% chance that 0.4 to 2.6 out of every 1000 items in the population will have "purple iris."  If you are strictly interested in the lower tail of the distribution, you could also calculate that there is a 95% chance that less 2.3 out of every 1000 items have "purple iris."  
