Given the following SDE:
$$ dS = \mu S dt + \sigma S dW $$
Why when we write this in discrete form:
$$\Delta S = S_{i+1} - S_i = \mu S_i \Delta t + \sigma S_i \phi \sqrt{\Delta t}$$
The indices on the RHS are all i's? i.e. there is no i+1 there.
I am having in mind modelling of some stochastic variable $x(t)$. Then we can write that $x(t+1) = x(t) + \eta(t+1)$. So that we have a component for previous cumulative sum value: $x(t)$ and we have current noise: $\eta(t+1)$
It's a convenient choice that we make.
You choose some discretization unit $\delta$, and obtain the recurrence (which is an approximation of the true SDE): $$S_{i+1}-S_i = \mu S^*\delta + \sigma S^*Z\sqrt{\delta},$$ where $Z$ is a standard normal random variable, and where $S^*$ is some point in the interval $[S_i,S_{i+1}]$. To obtain the most useful recurrence, we choose $S^*=S_i$, so $$S_{i+1} - S_i = \mu S_i\delta + \sigma S_i Z\sqrt{\delta},$$ because then $$S_{i+1} = S_i + \mu S_i\delta + \sigma S_i Z\sqrt{\delta},$$ and whenever you have $S_i$, you can readily obtain the next step $S_{i+1}$.
Because $S$ has no single value over an interval of the form $[t_i,t_i+\delta]$, you could have chosen any value of $S^*\in [S_i,S_{i+1}]$, but by the reasons above, one usually chooses $S^*=S_i$.