In Section 34 (page 445) of Billingsley's Probability and Measure (3rd ed.), he says, regarding conditional expectation, that since $E[X \| \mathscr{G}]$ is $\mathscr{G}$-measurable and integrable, $E[X \| \mathscr{G}](x)$ can in principle be calculated just from knowing for each $G \in \mathscr{G}$ whether or not $x \in G$.
Question 1: How can $E[X \| \mathscr{G}](x)$ be calculated just from knowing for each $G \in \mathscr{G}$ whether or not $x \in G$?
My only idea is based on what I'd do in a specific case: Let's suppose $X:\mathbb{R}\to \mathbb{R}$, $\mathscr{G} = \mathscr{B}$ is the Borel $\sigma$-algebra and $\mu$ is Lebesgue measure. If I had access to the knowledge of which $B \in \mathscr{B}$ contain $x$, then in order to compute $E[X \| \mathscr{B}](x)$, I'd search through $\mathscr{B}$ until I found a (non-degenerate) interval $B_{1}$ such that $x \in B_{1}$, at which point I'd compute $E[X|B_{1}] := \frac{1}{\mu(B_{1})} \int_{B_{1}} X \, d\mu$. Then I'd continue searching through $\mathscr{B}$ until I found another interval $B_{2}\subseteq B_{1}$ such that $x \in B_{2}$ and the diameter of $B_2$ is less than half of that of $B_{1}$, at which point I'd compute $E[X|B_{2}]$. Continuing this way, I'd have $B_{1}\supset B_{2} \supset \cdots$ with $\{x\}=\bigcap_{n} B_{n}$ and by the Lebesgue differentiation theorem (LDT),
$ \lim_{n\to \infty} E[X\| B_{n}] = E[X\|\mathscr{B}](x). $
Question 2: Can this procedure for computing $E[X \| \mathscr{B}]$ be generalized?
Of course, one problem with generalizing is that the LDT doesn't hold in general. It's also not clear how one would choose the $B_{n}$'s.
Update As @CarlMummert pointed out, it doesn't make sense to try to calculate the value of $E[X \| \mathscr{G}](x)$ for any $x$, since conditional expectation is only defined almost everywhere. Thus I am accepting @MichaelGreinecker's answer.