You say at first that $(X,\mathcal{O}_X)$ is a locally ringed space, but then talk about schemes afterwards. Note that there are many locally ringed spaces which are far from being schemes.
The most likely definition of quasicoherent sheaf on a locally ringed space is a sheaf $\mathcal{F}$ so that there is an open cover $\{U_i\}_{i\in I}$ of $X$ so that over each $U_i$, we have a (possibly infinite) presentation
$$ \mathcal{O}_{U_i}^{\oplus I}\to\mathcal{O}_{U_i}^{\oplus J}\to \mathcal{F}\to 0.$$
Notice that a locally free sheaf $\mathcal{E}$ is of course of this form as there is an open cover $\{U_i\}_{i\in I}$ of the space so that on each $U_i$
$$ 0\to \mathcal{O}_{U_i}^{\oplus J}\xrightarrow{\sim} \mathcal{F}\to 0.$$
In particular, locally free implies quasicoherent.
In the context of schemes, we can define being quasicoherent as being locally of the form $\widetilde{M}_i$ for $M_i$ an $A_i$-module with respect to an open cover $\{\operatorname{spec} A_i\}_{i\in I}$ of the scheme $X$. Then a locally free sheaf can be locally written as $\widetilde{A^{\oplus I}}=\widetilde{A}^{\oplus I}\cong \mathcal{O}_{\operatorname{spec}A_i}^{\oplus I}$ over $\operatorname{spec}A_i$. In particular, a locally free sheaf is a fortiori a quasicoherent sheaf.
Honestly, I can't figure out how the book wants us to explain this fact. Based on the mention of section functors and localizations I assume we are supposed to use the stuff in sections 2.13 and 2.14, so let's give this a shot to answer your question 1.
$\DeclareMathOperator{\QCoh}{QCoh}
\newcommand{\QA}{\QCoh(\mathbb A^2)}
\newcommand{\m}{\mathfrak m}
\newcommand{\QU}{\QCoh(U)}
\newcommand{\Id}{\mathrm{Id}}
\newcommand{\F}{\mathcal F}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Ext}{Ext}
\DeclareMathOperator{\Spec}{Spec}
$
Let $T$ be the full subcategory of $\QA$ consisting of sheaves set-theoretically supported at $\{0\}$. In terms of modules, these are modules such that every element is annihilated by a power of $\m$. It is clearly a Serre subcategory.
- We have $j^*\colon \QA\to \QU$. It sends $T$ to $0$, so by the universal property (Theorem 13.9), it factors through $\QA\to \QA/T\to \QU$. We claim the second functor is an equivalence. The inverse is the composition $\QU\xrightarrow{j_*} \QA\to \QA/T$. Going one way, $j^*j_*\cong \Id_{\QU}$. Going the other way, we have a natural transformation on $\QA$:
$$
\Id_{\QA}\to j_*j^*.
$$
I believe that all we need to see is that it becomes an isomorphism after modding out by $T$.
Let's just identify $\QA/T$ with $\QU$ from now on.
- $T$ is localizing, since $j^*\colon \QA\to \QU$ has a right adjoint, $j_*$. In terms of Section 2.14, $j_*$ is the "section functor". Proposition 14.7(6) tells us already that $j_*$ is fully faithful. Proposition 14.7(3) tells us that if $\F\in \QU$ and $M\in T$,
$$ \Hom_{\QA}(M,j_*\F)\cong \Hom_{\QU}(j^*M,j^*j_*\F) \cong \Hom_{\QU}(0,\F)=0.
$$
Proposition 14.7(3) tells us that $$\mathrm{Ext}^1_{\QA}(M,\F)=0.$$So this gives us one inclusion.
It remains to prove that if $M\in \QA$ is such that $\Hom(A/\m,M)=\Ext^1(A/\m,M)=0$, then $G$ is in the essential image of $j_*$. For this we use Theorem 14.8. It gives us an exact sequence:
$$
0\to \tau M\to M\xrightarrow{\eta} j_*j^*M\xrightarrow{\pi}M'\to 0.
$$
It further tells us that the first and last term are in $T$ and that $\eta$ is an essential map. We are going to heavily use that $A/\m$ generates $T$, in order to show that $\eta$ is an isomorphism.
First, being torsion, if $\tau M\neq 0$ has some submodule isomorphic to $A/\m$. This would mean that $M\supseteq \tau M$ does as well, which contradicts the assumption that $\Hom(A/\m,M)=0$. So $\tau M=0$.
Finally, if $M'\neq 0$ it also has a submodule isomorphic to $A/\m$. We then have a short exact sequence
$$
0 \to M\to \pi^{-1}(A/\m)\to A/\m\to 0
$$
Since $\Ext^1(A/\m,M)=0$ by assumption, this sequence splits, so $A/\m$ embeds into $\pi^{-1}(A/\m)\subseteq j_*j^*M$, into a submodule disjoint from $M$. This contradicts the fact that $\eta$ is essential. So $M'=0$, and $\eta$ is an isomorphism as desired. This answers question 1, modulo some details.
For question 2, I'm not sure what constitutes a satisfactory answer. Since you are looking at the stalk, you can assume that $X$ is affine, say $\Spec R$, and $x = V(f_1,\ldots ,f_n)$ for some $f_i\in R$. If we let $U_i = X\setminus V(f_i) = \Spec R[f_i^{-1}]$
$$
X\setminus x = \bigcup_i U_i
$$
So by the property of being a sheaf, for any sheaf $M$ on $X\setminus x$, we have that $\Gamma(X,j_*M)=\Gamma(X\setminus x,M)$ is the kernel of
$$
\bigoplus \Gamma(U_i,M) \longrightarrow \bigoplus \Gamma(U_i\cap U_j,M)
$$
And I believe that by the exactness of localization, tensoring the above map with the local ring at $x$ will give you the stalk as the kernel. In practice, how hard is this kernel to find? I have no clue. I can show that in the case of $\mathbb A^2$, $j_*\mathcal O_{\mathbb A^2\setminus 0} \cong \mathcal O_{\mathbb A^2}$.
Best Answer
Let $A$ be a discrete valuation ring (for example your $\operatorname{Spec} k[x]_{(x)}$), $\;X=\operatorname{Spec} (A)=\{\zeta, f\}$ the corresponding affine scheme ( $f$ the closed point , $\zeta $ the generic point ) and $U=\{\zeta\}$ the unique non empty and non full open subset of $X$.
An $ \mathcal{O}_X$-module $\mathcal{F}$ consists of an $A$-module $M (=\mathcal F(X))$, a $K=Frac(A)$-module $N(=\mathcal F(U))$ and an $A$-linear map $M\to N$ corresponding to the sheaf restriction.
[Note the amusing and unusual fact that every presheaf on $X$ is automatically a sheaf since $X$ has no non-trivial covering!]
These data automatically give rise to a canonical morphism of $K$-vector spaces $$F:M\otimes_AK\to N $$ Characterization of quasi-coherence
The sheaf $\mathcal F$ is quasi-coherent if and only if $F$ is bijective.
And now you can boast that you can describe all quasi-coherent sheaves on $X$ !