I'm trying to undestand why
$${Var}\left(\log\frac{\hat S(t)}{1-\hat S(t)}\right) \approx \frac{1}{\left(1-S(t)\right)^2}\sum_{i:Y_{(i)}\le t}\frac{d_i}{r_i(r_i-d_i)}$$
I know that by Greenwood formula: $${Var}\left({\hat S(t)}\right) = {\left(\hat S(t)\right)^2}\sum_{i:Y_{(i)}\le t}\frac{d_i}{r_i(r_i-d_i)}$$
So I know how to get to second part by I don't see why:
$$\log\left(\frac{\hat S(t)}{1-\hat S(t)}\right) \approx \frac{1}{\left(1-S(t)\right)^2}$$
Again with a non-linear analytic function $f$, we may try to expand it around the mean $\mu = E[X]$ and have
$$ Var[f(X)] \approx Var[f(\mu) + f'(\mu)(X - \mu)] = f'(\mu)^2 Var[X]$$
which we have seen this many times when we apply the delta method. Applying this approximation in your example,
$$ \begin{align} Var\left[\ln\frac {\hat{S}(t)} {1 - \hat{S}(t)}\right] &\approx \left(\frac {1} {E[\hat{S}(t)]} + \frac {1} {1 - E[\hat{S}(t)]}\right)^2 Var[\hat{S}(t)] \\ &= \frac {1} {E[\hat{S}(t)]^2(1 - E[\hat{S}(t)])^2} Var[\hat{S}(t)] \end{align}$$
Now to estimate this, we plug-in $\hat{S}(t)$ for $E[\hat{S}(t)]$ and estimate $Var[\hat{S}(t)]$ with the Greenwood formula. A little simplification will help you to obtain the estimate $\displaystyle \hat{Var}\left[\ln\frac {\hat{S}(t)} {1 - \hat{S}(t)}\right]$