I'm reading Mumford's book, precisely I'm looking at Proposition 1, page 28. In the image below I included the definition too, because it is necessary to give a sense to the proposition. Now, basically I don't understand what is $R_f $: is it a notation for $R_{(f)} $, where $(f) $ is the ideal generated by $f ?$ Moreover, I was convinced that $o_X (U) $ consisted of the functions $\frac f g$, with $f,g\in R $ and $g $ never vanishing on $U $; however it turns out that it is not true, and that the Proposition 1 is involved. So can you make clear to me what does the proposition mean? Thanks in advance
$\newcommand{uo}[]{\underline{o}}$
As discussed in the comments: $R_f$ is the localization $S^{-1}R$ with $S=\{1,f,f^2,f^3,\cdots\}$. Proposition 1 identifies the value of the sheaf $\uo_X$ on the open set $X_f$ to be $R_f$.
Let's try to work through your confusion about the value of $\underline{o}_X$ on open subsets not of the form $X_f$. On the one hand, if we have an irreducible variety $X$ with affine coordinate ring $R$ and structure sheaf $\uo_X$, then as Mumford notes, the rings $\uo_X(U)$ for every open set $U\subset X$ and every stalk $\uo_{X,x}$ inject in to $K=\operatorname{Frac} R$, so we can view elements of $\uo_X(U)$ or $\uo_{X,x}$ as elements of $K$. The problem comes when we try to understand what equality means for these images in $K$ - it's more subtle than "they're the same map $U\to k$".
Let us attempt to understand when two elements are equal in $K$. Suppose $\frac{f_1}{g_1}=\frac{f_2}{g_2}$ with $f_i,g_i\in R$. By the definition of the fraction field of an integral domain, these are the same iff $f_1g_2=f_2g_1$, which corresponds to the two functions agreeing on $D(g_1g_2)$. In particular, equality means that there's an open set on which the two functions eventually restrict to be equal on.
So it may happen that there are functions $F_1=\frac{f_1}{g_1}$ on $X_{g_1}$ and $F_2=\frac{f_2}{g_2}$ on $X_{g_2}$ which are equal on $X_{g_1g_2}$ (and thus equal as elements of $K$) and patch together to form a function on $X_{g_1}\cup X_{g_2}$, but without the formula $\frac{f_1}{g_1}$ or $\frac{f_2}{g_2}$ being valid on the whole of $X_{g_1}\cup X_{g_2}$.
The example Mumford gives is instructive. Let $X=V(xy-zw)\subset k^4$, with $R=k[x,y,z,w]/(xy-zw)$. As $xy-zw$ is irreducible, we have that $X$ is an irreducible algebraic set, with fraction field $K=\operatorname{Frac} R$. Now consider the two formulas $\frac{x}{w}$ and $\frac{z}{y}$. The first defines a function on $X_w$, and the second defines a function on $X_y$. These are equal on the overlap $X_{wy}$, as $\frac xw=\frac zy$ there, so they glue to a function on $X_{w}\cup X_y$ and they're also equal as elements of $K$. But there is no one formula in $K$ that will define this function on $X_w\cup X_y$: a specific formula $f/g$ will always be defined on $X$ away from $V(g)$, and $V(g)\subset X$ is either empty, codimension one, or the whole space (for irreducible affine $X$). By contrast, the function on $X_{w}\cup X_{y}$ is defined on the complement of a set of codimension two, and there's no way to extend the function over this. So there is no one choice of a formula for the function we constructed which can be valid on the whole of $X_w\cup X_y$.
So what has happened here? The format we've chosen to write down representatives of our functions has failed us, because the version of equality we have inside $K$ isn't the same as what we need to write down functions in general. What might be better said here is that $\uo_X(U)=\cap_{x\in U} \uo_x$ gives us representatives of functions that can be extended to the whole of $U$, not that the elements of $\uo_X(U)$ necessarily define functions which are valid on the whole of $U$. Proposition 1 tells us that these two concepts are actually the same when $U$ is the nonvanishing set of a regular function $f$ on $X$.