The question asks how we go from $\displaystyle p=1 - \frac{N!}{(N-k)!\,N^k}$ for the probability of collision of $k$ uniformly random values among $N$, to the approximation $\displaystyle p\approx\frac{k(k-1)}{2N}$ (which assumes $k$ is small compared to $\sqrt N$ ).
First we get back to $\displaystyle1-p=\prod_{j=0}^{k-1}{\left(1-\frac j N\right)}$, which is how $p$ was determined in the first place. Then we take the logarithm on both sides and use that $u>0,v>0\implies\ln(u\,v)=\ln(u)+\ln(v)$ to get
$$\displaystyle\ln(1-p)=\sum_{j=0}^{k-1}{\ln\left(1-\frac j N\right)}$$
For small $|x|$, it hold $\ln(1+x)\approx x$. Applying this to $x=p$ on the left side and $\displaystyle x=\frac j N$ on the right side, we get $\displaystyle p\approx\sum_{j=0}^{k-1}\frac j N$. We rewrite this as $\displaystyle p\approx\frac 1 N\sum_{j=0}^{k-1}j$.
Now we use that the sum of non-negative integers less than $k$ is $\displaystyle\frac{k\,(k-1)}2$ and get the desired $\displaystyle p\approx\frac{k(k-1)}{2N}$.
Without proof: this approximation of $p$ is always by excess. It's off by less than $+28\%$ when $k\le\sqrt N$, less than $+14\%$ when $k\le\sqrt{2N}$, less than $+7\%$ when $k\le2\sqrt N$.
Most of the error is from the approximation $\ln(1-p)\approx-p$. A much better approximation is:
$$p\approx1-e^{-\frac{k(k-1)}{2N}}$$
which assumes only $k\ll N$ rather than $k\ll\sqrt N$. However beware that this alternate formula is numerically instable for small $p$.
In comment it's additionally asked
How shall I understand (this formula) from $1/N$ for each pair? Are the pairs each having two equal values disjoint events? Which part in its derivation is approximation?
One easy way to derive the probability $p$ that there is a collision among $k$ uniform values among $N$ (for $0\le k\le N$) is as the complement of the probability that there is not collision.
For a fixed $N$, define $q_k$ as the probability that there is no collision after $k$ values. Obviously $q_0=q_1=1$. And for $k\ge2$, $q_k$ is the probability that there was no collision among the first $k-1$ values (that is, $q_{k-1}$), time the probability that there is no collision between the $k-1$ previous values and the last drawn one, which is $\displaystyle\frac{N-k}N$ (justification there a exactly $N-k$ values among $N$ that do not leak to collision for the last value drawn).
It follows $\displaystyle q_k=q_{k-1}\left(1-\frac k N\right)$, thus $\displaystyle q_k=\prod_{j=0}^{k-1}{\left(1-\frac j N\right)}$, thus $$p=1-\prod_{j=0}^{k-1}{\left(1-\frac j N\right)}=1-\frac{N!}{(N-k)!\,N^k}$$
This is exact. See first two sections of this answer for the derivation of approximations.
One source justified the approximation as:
One way of looking at this is that if you choose $k$ elements, then there are $k(k−1)/2$ pairs of elements, each of which has a $1/N$ chance of being a pair of equal values.
This hand-waving argument does not yield a mathematically exact derivation of $\displaystyle p=\frac{k(k-1)}{2N}$, since the events are not disjoint. As long as $p$ is small, we can get away with it, but that gets grossly wrong when $k>\sqrt{2N}$.
When $k = \sqrt{N}$, this chance is close to 50%.
That's true if 39.3% is close to 50%.
