Neither of these claims are true, and the reason is essentially that "The topology on $X\times Y$ is not the product topology". This is more or less the claim you're making in part $2)$, and then this failure causes $1)$ to fail. The fact that the topology isn't just the product topology is maybe not entirely surprising at the level of schemes (since the product itself isn't even the product as sets), but I think it is relatively surprising if you're thinking about this on the level of classical varieties, where you can just define the set of the product as the usual product, but then have to work harder to define the topology. This difference is because one of the main advantages of using schemes rather than classical varieties is that the points of a scheme contain a lot of information about the topology then the points of the classical variety (which are just the closed points of the scheme).
An example where $2)$ fails is where $X = Y = \mathbb{A}^1_k$, with $k = \bar{k}$ for simplicity. Then any non-empty open set in $\mathbb{A}^1_k$ is just the complement of a finite set of (closed) points. Explicitly, any open set is of the form $D(f)$ for some $f$ and corresponds to the complement of the roots of $f$. Then the product $D(f) \times_k D(g) = \pi_X^{-1}(D(f)) \cap \pi_Y^{-1}(D(g))$ corresponds to the complement of a finite union of lines in $\mathbb{A}^2_k$, corresponding to vertical lines corresponding to the roots of $f$ and horizontal lines corresponding to the roots of $g$. Note that a union of sets of this form is still of this form, so the topology generated by these sets is simply their union. In particular, there are many open sets not of is form, e.g. the diagonal $D(X-Y)$, so the actual topology is much richer than the product topology.
Moreover, $1)$ is also not true. You could probably just compute the product of the preimage sheaves in the example above, but there is an easier way that fits in with the philosophy used above that the product of schemes is more than just the naïve product. As well as the topology not being the product topology, the set is not even the product of the sets! In particular, there may be distinct points in the product that project to the same points in $X$ and $Y$. To see that this is a problem, note that:
$$(\pi^{-1}_X \mathcal{O}_X \otimes \pi^{-1}_Y\mathcal{O}_Y)_z \cong (\pi^{-1}_X \mathcal{O}_X)_z \otimes_k (\pi^{-1}_Y\mathcal{O}_Y)_z \cong \mathcal{O}_{X, \pi_X(z)} \otimes_k \mathcal{O}_{Y,\pi_Y(z)} $$
And so in particular, points with the same projections to $X$ and $Y$ should have the same stalks. However, with our example above, notice that the primes $(X-Y)$ and $(0)$ both map to the generic point under either projection to $\mathbb{A}^1$, and so the stalks of the product sheaf at both points are the same. However, the stalks of the structure sheaf are different (and we can even see this without calculation, the local rings have different dimensions since the closures of the two points have different codimension in $\mathbb{A}^2_k$).
As Paf points out in the comments to your question, you can look at this post on mathoverflow for a little more discussion on what the points of the fibre product are, and what stalks of the structure sheaf of the product are at said points. In particular there are a wealth of points that where the product sheaf will have the same stalks, but the structure sheaf won't. Notice that the same argument will extend to show that we can't even replace the tensor product of the preimage sheaves with the tensor product of the pullback sheaves, or any other reasonable extension.
My question: Is there any relation in general between the stalk $\mathcal O_{X_y,x}$ and $\mathcal O_{X,x}$?
Yes. If $\mathfrak m_y$ denotes the maximal ideal of $\mathcal O_{Y,y}$, then $$\mathcal O_{X_y,x}\cong \mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}\,,$$
where $\mathfrak m_y\mathcal O_{X,x}\subseteq \mathcal O_{X,x}$ denotes the ideal generated by the image of $\mathfrak m_y$ under the ring morphism $\mathcal O_{Y,y}\rightarrow \mathcal O_{X,x}$. To see this, we may assume $X\cong\operatorname{Spec} B$ and $Y\cong\operatorname{Spec} A$ are affine, with $x$ and $y$ corresponding to prime ideals $\mathfrak q\subseteq B$ and $\mathfrak p\subseteq A$. Then $X_y\cong \operatorname{Spec}(B\otimes_Ak(\mathfrak p))\cong\operatorname{Spec}B_{\mathfrak p}/\mathfrak p B_{\mathfrak p}$ (here $B_{\mathfrak p}$ denotes the localisation of the $A$-module $B$ at $\mathfrak p$; moreover, $\mathfrak p B_{\mathfrak p}\subseteq B_{\mathfrak p}$ denotes the ideal generated by the image of $\mathfrak p$). Thus the local ring $\mathcal O_{X_y,x}$ is given by the localisation of the ring $B_{\mathfrak p}/\mathfrak p B_{\mathfrak p}$ at its prime ideal $\mathfrak q/\mathfrak pB_{\mathfrak p}$. By elementary properties of localisation, this coincides with $B_{\mathfrak q}/\mathfrak p B_{\mathfrak q}\cong \mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}$.
Also note that $\mathcal O_{X_y,x}$ has the same residue field as $\mathcal O_{X,x}$. Indeed, under the isomorphism $\mathcal O_{X_y,x}\cong \mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}$, the maximal ideal of $\mathcal O_{X_y,x}$ corresponds to $\mathfrak m_x/\mathfrak m_y\mathcal O_{X,x}$, and $$(\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x})/(\mathfrak m_x/\mathfrak m_y\mathcal O_{X,x})\cong \mathcal O_{X,x}/\mathfrak m_x\cong k(x)\,.$$
The author says: 'Indeed, the quotient $\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}$ remains unaltered when we pass from $X$ to the fiber $X_y$'. I'm confused about the meaning of this statement.
This is somewhat sloppy language and I'll try to rephrase it in a super pedantic way. Let $Y'=\operatorname{Spec} k(y)$ and let $X'=X\times_YY'\cong X_y$. Let $x'\in X'$ denote the unique point lying over $x\in X$ and let $y'\in Y'$ be the unique point (which automatically lies over $y$). Then what the author is trying to say is that $$\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}\cong \mathcal O_{X',x'}/\mathfrak m_{y'}\mathcal O_{X',x'}\,.$$
Indeed, the left-hand side equals $\mathcal O_{X',x'}$, as seen above, and on the right-hand side we have $\mathfrak m_{y'}=0$ since $Y'=\operatorname{Spec} k(y)$ is the spectrum of a field.
For the sake of completeness, let me elaborate a bit more on Liu's proof, since I think quite some things get swept under the rug. Assume first that all fibres $X_y$ are finite over the respective residue field $k(y)$. By the structure theory of artinian rings (see [Stacks Project, Tag 00J4]), $X_y$ has only finitely many points $x_1,\dotsc,x_n$, and $X_y\cong \coprod_{i=1}^n\operatorname{Spec} \mathcal O_{X_y,x_i}$.
If $X_y$ is reduced, then each of the artinian local rings $\mathcal O_{X_y,x_i}$ must be reduced. But a reduced artinian local ring must be a field, whence $\mathcal O_{X_y,x_i}\cong k(x_i)$. By the discussion above, this also implies $\mathcal O_{X,x_i}/\mathfrak m_y\mathcal O_{X,x_i}\cong k(x_i)$. In particular, if each $k(x_i)$ is separable over $k(y)$, then $f$ is indeed unramified.
Conversely, assume $f$ is unramified. We must show that each fibre $X_y$ is finite and reduced over $k(y)$ (and that all residue fields $k(x)$ are separable over $k(y)$, but this is automatic from the assumption that $f$ is unramified). Since $f$ is unramified, $\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}\cong k(x)$ holds for all $x\in X_y$. Thus, all local rings $\mathcal O_{X_y,x}\cong k(x)$ are fields. In particular, they are all 0-dimensional, which implies that $X_y$ must be zero-dimensional. This implies again that $X_y$ has only finitely many points $x_1,\dotsc,x_n$ and that $X_y\cong \coprod_{i=1}^n\operatorname{Spec} \mathcal O_{X_y,x_i}$ (see [Stacks Project, Tag 0AAX]; the disjoint union must be finite since $X_y$ is quasi-compact). Since all $\mathcal O_{X_y,x_i}\cong k(x_i)$ are reduced and finite-dimensional over $k(y)$, we see that $X_y$ too must be reduced and finite over $k(y)$, as claimed.
Best Answer
Fiber products exist in the category of locally ringed spaces (see e.g. Gillam's paper), and this also provides a direct construction (without gluing!) of the fiber product of schemes and reveals its explicit structure as a locally ringed space.
If $f : X \to S$ and $g : Y \to S$ are morphisms of locally ringed spaces, then the fiber product $X \times_S Y$ has the following description: Elements are of the form $(x,y,s,\mathfrak{p})$, where $(x,y,s)$ lies in the underlying topological fiber product, i.e. $x \in X$, $y \in Y$ with $f(x)=s=g(y)$, and $\mathfrak{p} \subseteq \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y}$ is a prime ideal satisfying $\mathfrak{p} \cap \mathcal{O}_{X,x} = \mathfrak{m}_x$ and $\mathfrak{p} \cap \mathcal{O}_{Y,y} = \mathcal{m}_y$. The stalk of the structure sheaf at such a point is the localization $(\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y})_{\mathfrak{p}}$. As for the topology and the structure sheaf in general, one uses (as in the construction of affine schemes) localization at elements $f \notin \mathfrak{p}$.
Actually, all this can be derived from the universal property of local schemes and the universal property of fiber products (exercise).