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If I wish to use finite difference methods to approximate the pricing function $F(t, s)$ for an option (say, a call), what size grid should I use?
I mean, it seems to make sense to start the grid at zero for both variables $t, s = 0$, and then let the upper bound on the $t$-grid be $T$ (the maturity of the option)... is this true?
And what about the upper bound on $s$?
For the maturity, choose a grid $\{t_0=0,t_1,\dots,t_n=T\}$ such that $T$ is the option's maturity.
For the underlying, if it is positive, you might choose an upper boundary by selecting a grid $\{S_0=0,S_1,\dots,S_{\max}\}$ such that the derivative's delta at $t_{n-1}$ is above a threshold $D$ in order to specify a boundary condition such as:
$$ \frac{\partial V}{\partial S}=1, \quad \frac{\partial^2 V}{\partial S^2}=0 $$
You can determine the upper value $S_{\max}$ by starting to compute the derivative's value at $t_{n-1}$ for $S_0$ then proceeding up until:
$$ \frac{\partial V}{\partial S}>D $$
Answer is very straightforward: always center your grid on the current spot value and make sure that it covers sufficiently many standard deviations on each side to cover enough of the terminal distribution at maturity. In particular it should cover the forward at maturity. In practice if you use something like 5stdev on each side should be sufficient for most purposes
The other answer is sadly wrong. It doesn't make sense to say that you should "center your grid on the current spot value", because by definition of you trying to approximate $F(s)$ for all reasonable $s$, there IS no "current spot value". You are trying to find the price for all of them, right? So why should you center it around some imaginary current value?
That'd make sense if you wanted the price for, say, F(s = 100), specifically.
A good rule of thumb: 2-5 times the strike is your upper bound on the spots.
