Interpretation of an equation I'm trying to describe the equation in the attached screenshot in my own words. 
$\lvert(\overline\chi_1-\overline\chi_2)\rvert > s_1(t_{1 @ 99 \%}) + s_2(t_{2 @ 75 \%})$
The equation tests whether 99% of population 1 lies outside of 75% of population 2. The equation is taken from Donegan, T., 2012. Geographical variation in Immaculate Antbird Myrmeciza immaculata, with a new subspecies from the Central Andes of Colombia. Bull. Brit. Orn. Club, 132, pp.1–40.
My current interpretation is as follows: 
The difference between the means of the population 1 and 2 must be greater than 99% of a standard deviation in population 1 plus 75% of a standard deviation in population 2. 
I realise my interpretation may not be quite right. Can anybody give a better interpretation?
 A: This is a comment and a partial answer.  I hope it might prompt ways to clarify the question (and if it does, I will be happy to delete it).
We need to make some assumptions based on clues in the question and typical statistical terminology, but let me make them fully explicit:


*

*$\bar{\chi}_1$ and $\bar{\chi}_2$ are means of something.

*$s_1$ and $s_2$ are the corresponding standard deviations of something.

*"$t_{df @ q\%}$" is idiosyncratic notation, but likely it refers to the upper $1-(1-q)/2$ quantile of Student's t distribution having $df$ degrees of freedom.  Thus, $t_{1 @ 99\%}$ refers to the $0.995$ quantile of the t distribution with either $1$ df (or else the "$1$" denotes the degrees of freedom inherent in estimates from a sample of population 1) and $t_{2 @ 75\%}$ refers to the $0.875$ quantile of the t distribution with $2$ df (or maybe the $2$ references population 2).
The appearance of the Student t suggests that the $\bar{\chi_i}$ are actually sample means, that the $s_i$ are standard errors of the mean (not standard deviations!), and that the underlying population is assumed to have a distribution that is not too skewed.  Let the population means be $\mu_1$ and $\mu_2$.  These assumptions imply that the $X_i = (\bar{\chi}_i - \mu_i) / s_i$ have (approximately) a Student t distributions.
Please be aware of the sharp and important distinction between a sample standard deviation and a standard error of the mean: with a sample of $n$ values, the former (which estimates the population standard deviation) is $\sqrt{n}$ times the latter.
I have deduced that the quantile should be $1-(1-q)/2$ rather than $q$ itself because it is not stipulated beforehand whether $\mu_1$ is greater or less than $\mu_2$.
Now we can do some algebra with what we have in order to recreate the expression in the question:
$$|\bar{\chi}_1 - \bar{\chi_2}|  = |s_1 X_1 + \mu_1 - (s_2 X_2 + \mu_2)| = |s_1 X_1 - s_2 X_2 + (\mu_1-\mu_2)|$$
where the $X_i$ have Student t distributions.
That's as far as we can get without more information: the right hand side is not directly related to $s_1 t_{1@99\%} + s_2 t_{2@75\%}$.
Now, it is true that $s_1 t_{1@99\%} + s_2 t_{2@75\%}$ is a sum of multiples of two standard errors (or perhaps standard deviations).  However, it is quite unclear what this sum would be testing, because it is not directly related to any probability.  It looks like a corrupted (and invalid) version of a two-sample t-test.
If we instead interpret the $s_i$ to be sample standard deviations, then the right hand side in the question could be the result of the following (erroneous) reasoning:

The data in population 1 have a standard deviation of $s_1$ so we assume that population is distributed around its mean as $s_1$ times a Student t distribution; and likewise for population 2. Thus if the means of those distributions are separated by more than (1) the distance from the mean within which 99% of all data in population 1 lie plus (2) the distance from the mean within which 75% of all data in population 2 lie, we may conclude that "99% of population1  lies outside of 75% of population 2."

Unfortunately, extremely few populations actually have a Student distribution and there appears to be no theoretical basis to introduce the Student t distribution in such a context.
The idiosyncratic notation, absence of any reference to the true population means $\mu_i$, and lack of a natural interpretation of the right hand side of the expression in the question, all suggest there's a good likelihood this expression is erroneous.  Without more information about what the notation is intended to mean, what the data are, and what the test is, it is difficult to tell for sure.
A: Edit - just read @whuber's fantastic answer on this and agree with what he has said but I've left mine as further illustration of his second possibility.  I fully agree with him that

Unfortunately, extremely few populations actually have a Student
  distribution and there appears to be no theoretical basis to introduce
  the Student t distribution in such a context.


My original answer
I think it actually probably means that the right hand side is the standard deviation of the first population times the 99th percentile of a t distribution relevant to that population, + the standard deviation of the second population times the 75th percentile of a t distribution relevant to that population.
This only makes sense if for some reason the population is being modelled with a shifted t distribution.
Consider the plot below which shows two such distributions.  The blue vertical line has 75% of population2 on its right; the red vertical line has 99% of population1 on its left.  This comes from a simulated population where the mean of population 1 is 8, that of population 2 is 30, and the standard deviations are 5 and 8.

s1 <- 5
s2 <- 8
mean1 <-8
mean2 <- 30

density100 <- dt(seq(from=-3, to=3, length=100), 15)

plot(seq(from=mean1-3*s1, to=mean1+3*s1, length=100), density100, 
    type="l", col="red", xlim=c(-10,50), bty="l")
lines(seq(from=mean2-3*s2, to=mean2+3*s2, length=100), density100, 
    type="l", lty=2, col="blue")

abline(v=mean1+s1*qt(.99,df=15), col="red")
abline(v=mean2-s1*qt(.75,df=15), col="blue")

text(34.5,.1, "75% of a shifted t\ndistribution is in here", cex=.7, col="blue")
text(6,.1, "99% of a shifted t\ndistribution is in here", cex=.7, col="red")

qt(.99, df=15)*s1 + qt(.75, df=15)*s2

which gives:
> qt(.99, df=15)*s1 + qt(.75, df=15)*s2
[1] 18.54198

This is less than the difference between the two means (30-8=22 > 18.5).
