Why does the following function distribute things in a binomial distribution?
The second function is easier to explain: it sends tract $t$ to position $t + h \pmod{n}$, where $h = hash(g)$. The function $t \mapsto t + h \pmod{n}$ is injective (one-to-one), and so every index $i$ gets mapped exactly once (in fact, by $i - h \pmod{n}$).

For the first function, we can think of $t \mapsto hash(g+t)$ as a random function (this is a common assumption whenever hash functions are involved; the fact that we use SHA-1 is completely irrelevant). Therefore for each fixed index $i$, the probability that $t$ is mapped to $i$ is $1/n$, and the total number of $t$s mapped to $i$ is distributed $\mathrm{Bin}(1/n,1) \approx \mathrm{Po}(1)$, that is it has binomial distribution which is approximated well by a Poisson distribution with mean $1$.