The abstract mathematical machinery is intuitive for many people provided it is clearly set up in a way that parallels the original probability problem.
Before reading any more of this post, though, please skip to the end: if the argument made there is anything but intuitive to you, then you can return here and read the details.
Analysis
The idea is to create an obviously correct mathematical model of the draws by mechanically and directly representing those draws with mathematical objects. At this stage of the analysis we will minimize the amount of thought and insight needed, so we can be sure of capturing any subtleties that might elude our intuition.
The first step is to represent the experimental outcomes accurately and completely. The first draw will be one of the $n = r+g$ balls; the second draw will be a different one of the balls. Thus, if we were to number the balls $1, 2, \ldots, n$ to distinguish them, then the possible draws are represented by all ordered pairs of indices $(i,j)$ where $1\le i\le n,$ $1\le j \le n,$ and $i\ne j.$ The first entry in an ordered pair designates the first ball drawn and the second entry designates the second ball drawn.
This is the sample space $X.$
The second step is to determine the probabilities. You know two things about them:
All balls are equally likely to be the first draw. Thus, every subset of the form $$\mathcal E_i = \{(i,1), (i,2),\ldots,(i,i-1),\quad (i,i+1),\ldots, (i, n)\}$$ has the same probability. Since there are $n$ such events, $$\mathbb P(\mathcal E_i) = \frac{1}{n}$$ for every $i.$
All remaining balls are equally likely to be the second draw. This says that within each $\mathcal E_i,$ each of the $n-1$ ordered pairs is equally likely. Allocating the total probability $\mathbb P(\mathcal E_i)$ to each of its members thereby gives each of those members a probability $$\mathbb P(\{(i,j)\})=\frac{1}{n-1}\times\mathbb P(\mathcal E_i) = \frac{1}{n(n-1)}.$$
We have thereby constructed a probability space out of this sample space and endowed it with a probability measure $\mathbb P$ that models the problem. This space has the salient characteristics that
All elements of $X$ have probabilities and
Those probabilities are all equal (to $1/(n(n-1))$): all singletons are equiprobable.
(Having found this result, we could have argued at the outset that all pairs are equally probable--and thereby avoid doing any algebra at all, as you will see.)
As is often the case with finite sample spaces, the solution to any question of probabilities often devolves to counting. These events are:
Let $G$ be the subset of indices in $\{1,2,\ldots, n\}$ corresponding to the green balls. $G$ therefore has $g$ elements. Use it to express the first event explicitly:
$$\mathcal G_1 = \{(i,j)\mid i\in G,\ 1\le i\le n,\ 1 \le j\le n,\ i\ne j\}.$$
The second event is expressed in the same way, merely with the roles of $i$ and $j$ switched:
$$\mathcal G_2 = \{(i,j)\mid j\in G,\ 1\le i\le n,\ 1 \le j\le n,\ i\ne j\}.$$
Since the question asks us to see why two events have the same chances, we would be well served by some insightful observation rather than mechanical counting or applying some combinatorial formula. This motivates studying the permutation $\pi:X\to X$ given by swapping the indices:
$$\pi((i,j)) = (j,i)$$
for all $(i,j)\in X.$
Here, then, is the full rigorous argument:
$\pi$ does not change the probability of any event, since every $(i,j)$ is equiprobable. Because $\pi(\mathcal G_1)=\mathcal G_2$ and $\pi(\mathcal G_2) = \mathcal G_1,$ $\mathcal G_1$ and $\mathcal G_2$ have the same probabilities.
If you care to write this formally (which is one method mathematics uses to develop insight) it might go like this:
$$\begin{aligned}
\mathbb P(\mathcal G_1) &= \sum_{i\in G}\sum_{j\ne i} \mathbb P(i,j)&\text{Axiom}_1\\
&= \sum_{i\in G}\sum_{j\ne i} \mathbb P(\pi(i,j))&\pi\text{ preserves }\mathbb P \\
&= \sum_{i\in G}\sum_{j\ne i} \mathbb P(j,i)&\text{Definition of }\pi \\
&= \sum_{j\in G}\sum_{i\ne j} \mathbb P{(i,j)}&\text{Relabel the subscripts}\\
&= \mathbb P(\mathcal G_2).&\text{Axiom}_1
\end{aligned}$$
"$\text{Axiom}_1$" is the probability axiom that the probability of a finite union of disjoint events is the sum of the probabilities of those events.
Solution
By reflecting on this demonstration and the insight afforded by $\pi,$ you might want to capture the intuition in words. Here's one effort:
Because there are just as many pairs of draws in which the first ball is green as there are pairs of draws in which the second ball is green, and every pair of draws has the same chance of occurring, the probability the first ball is green equals the probability the second ball is green.
As with many mathematical discoveries, in retrospect the result looks obvious. That's what is meant by "intuitive."
