I recently started studying the theory of schemes from "The geometry of schemes", by Eisenbud and Harris. At the very beginning of this volume I unfortunately found the first osbtacle: pick a commutative ring with unity $R$, and consider $f \in R$ as a function on Spec$R$. The authors define $f(\mathfrak{p})$ as the image of $f$ via the canonical maps $$R \rightarrow R/\mathfrak{p}\rightarrow k(\mathfrak{p}) $$ with $k(\mathfrak{p})$ the residue field of $R/\mathfrak{p}$.
If I'm not wrong, $k(\mathfrak{p})=((A/\mathfrak{p})\setminus \{[0]\})^{-1}A/\mathfrak{p}$, which should be equal to $k(\mathfrak{p}) \simeq A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ (but I can't figure out how, all this localisations confuse me).
The problem is that I'm getting confused with all this stuff, I mean, I always naively consider the function $15$ at $(7)$, then $15((7))=1$ mod $7$, or working over $k[x]$, $$x^3-1((x^2+1))= -x-1 (\text{ mod } (x^2+1)).$$
But I don't see all this machinery behind this computation, I can do the calculation but I feel I don't undertand what $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ is and why I'm working in this exact space.
Any help which clarifies the theoretic aspect of this calulations would be appreciate, I'm quite confused about this ring. Thanks in advance.
Maybe it helps to put in something more geometric. Let's say $k = \bar{k}$ is an algebraically closed field and you look at something like $\mathbb A^n := \mathrm{Spec} \ k[X_1,...,X_n]$.
Let's first consider a global section $f = \sum_{\underline i} a_{\underline i} X_1^{i_1}\cdot ... \cdot X_n^{i_n} \in \Gamma(\mathbb A^n, \mathcal O_{\mathbb A^n}) = k[X_1,...,X_n]$.
Take any closed point $\mathfrak{m} \in \mathbb A^n$. By Hilbert's Nullstellensatz we know that there exist $x_1,...,x_n \in k$ such that $\mathfrak{m} = (X_1 -x_1, ...., X_n-x_n)$.
Now observe that the residue field is given by $\kappa(\mathfrak m) = k[X_1,...,X_n]/\mathfrak m \cong k$ where the last isormophism is given by $X_i \mapsto x_i$.
Now we can compute:
$f(\mathfrak m) = \sum_{\underline i} a_{\underline i} X_1^{i_1}\cdot ... \cdot X_n^{i_n} \equiv \sum_{\underline i} a_{\underline i} x_1^{i_1}\cdot ... \cdot x_n^{i_n} \mod \mathfrak m = f(x_1,...,x_n) \in k$. Hence if we think of $\mathfrak m$ as the point of $\mathbb A^n$ with coordinates $(x_1,...,x_n)$, this definition of evaluation makes perfectly sense.
Also note that it makes sense that for example $\Gamma(D(g),\mathcal O_{\mathbb A^n}) = k[X_1,...,X_n]_g$:
A function on $D(g)$ is of the form $\frac{f}{g^k}$.
When is a closed point $(x_1,...,x_n)$ in $D(g)$? Well, $\mathrm{Spec} \ k[X_1,...,X_n]_g \cong \{\mathfrak p \in \mathrm{Spec} \ k[X_1,...,X_n] \mid g \not\in \mathfrak p\}$.
What does it mean of a maximal ideal $\mathfrak m := (X_1-x_1,...,X_n-x_n)$ to contain $g$?
$g \in \mathfrak m \iff g \equiv 0 \mod \mathfrak m \iff g(x_1,...,x_n) = 0$.
Hence: If $(x_1,...,x_n) \in D(g)$, then $g(x_1,...,x_n) \neq 0$ and hence $\frac{f(x_1,...,x_n)}{g(x_1,...,x_n)} \in k$ is well-defined (and you can check that this really is the evaluation you get using your definition above).
Note that we have non-closed points $\mathfrak p$ in $\mathbb A^n$ as well. For example the generic point $(0)$. The evaluation at these points won't give us actual numbers, but something different instead. For example $f((0)) = f \in k(X_1,...,X_n)$.
Working over non-algebraically closed fields we might end up getting different residue fields, even in the very "geometric" example of $\mathbb A^n_k$. For example if $n = 1$ and $k = \mathbb Q$ we have the closed point $(X^2 - 2)$ so that $\kappa( (X^2 - 2) ) = \mathbb Q(\sqrt 2)$.
In the case of $\mathrm{Spec} \ \mathbb Z$ the situation is even more weird:
The residue field at the generic point $(0)$ is given by $\mathbb Q$ and the residue field at a prime $(p)$ is given by $\mathbb F_p$. Hence a function $x \in \Gamma(\mathrm{Spec} \ \mathbb Z, \mathcal O_{\mathbb Z}) = \mathbb Z$ takes values in different fields at every point.
However, it still makes sense to evaluate at every point. And the same intution works for sets like $D(2) = \{(2)\}^{c}.$ As on a point $\mathfrak p \in D(2)$, one has that $2 \in \kappa(\mathfrak p)^\times$, it makes sense to compute $\frac{f}{2^k}(\mathfrak p) = \frac{f(\mathfrak p)}{2(\mathfrak p)} \kappa(\mathfrak p)$.
// Edit: Also, regarding your question about the residue fields: Let $A$ be a ring, $S$ a multiplicative subset and $I \subseteq A$ an ideal.
Take the exact sequence $0 \to I \to A \to A/I \to 0$ of $A$-modules.
Applying the exact functor $S^{-1}$ yields the exact sequence:
$0 \to S^{-1}I \to S^{-1}A \to S^{-1}(A/I) \to 0$ of $S^{-1}A$-modules, hence also of $A$-modules via the canonical map $A \to S^{-1}A$.
That is: $S^{-1}A/I \cong S^{-1}A / S^{-1}I$ as $A$-modules.
Both sides are rings, and it is not too difficult to check that the above map sends $1 \mapsto 1$ and that it is multiplicative, hence a ring isomorphism.
This gives you the desired isomorphism that you described in your question.
PS: Ravi Vakil's notes try to be very geometric and are filled with exercises that try to explain, why Algebraic Geometry deserves to be called Geometry.