One way is to show that $M \otimes N$ modulo the the submodule $P$ generated by $m' \otimes n$ and $m \otimes n'$ (with $m' \in M', m \in M, n' \in N', n \in N$) directly satisfies the same universal property as $M/M' \otimes N/N'$.
You have a well defined $A$-bilinear map $j: M/M' \times N/N' \rightarrow (M \otimes N)/P$ given by
$$(m+M',n+N') \mapsto m \otimes n + P$$
Suppose $Q$ is an $A$-module, and $f: M/M' \times N/N' \rightarrow Q$ is $A$-bilinear. I claim that there exists a unique $A$-module homomorphism $\phi: (M\otimes N)/P \rightarrow Q$ such that $\phi \circ j = f$. The uniqueness of the tensor product will then guarantee you an isomorphism $M/M' \otimes N/N' \rightarrow (M \otimes N)/P$ such that $(m +M') \otimes (n+N') \mapsto m \otimes n + P$.
The map $M \times N \rightarrow Q, (m,n) \mapsto f(m + M',n + N')$ is certainly $A$-bilinear, so there exists a unique $A$-linear map $\psi: M \otimes N \rightarrow Q$ given on generators by $\psi(m\otimes n) = f(m+M',n+N')$. Also, for any $m' \in M', n' \in N$ you have
$$\psi(m' \otimes n) = f(m' + M',n + N') = f(0 + M', n+ N') = 0$$
$$\psi(m \otimes n') = f(m + M', n' + N') = f(m+M',0+N') = 0$$
so $P$ is contained in the kernel of $\psi$. Therefore the $A$-module homomorphism $\phi: (M \otimes N)/P \rightarrow Q$ defined by $\phi(x + P) = \psi(x)$ is well defined.
Now $\phi$ does what is required:
$$\phi \circ j(m+M',n+N') = \phi(m\otimes n + P) = \psi(m \otimes n) = j(m+M',n+N')$$
and the uniqueness of $\phi$ is easily seen from the uniqueness of $\psi$.
Maybe it is better to see the construction this way: a map from $\operatorname{Spec}(K)$ to $X_k$ gives a
closed point $x$ of $X_K$. The underlying topological space of $\operatorname{Spec}(O_K)$ has two important points: a generic point coming from $\operatorname{Spec} K$ and a closed point $s$. So if we want to extend the map from $\operatorname{Spec}(K)\to X_K$ to a map $\operatorname{Spec}(O_K)\to X$ we have to send the generic point of $\operatorname{Spec}(O_K)$ to $x$ and because we want the continuity on topological spaces we have to send $s$ to a point in the closure of $x$ for example $z$. so topologically the image of our map must be in $Spec(O_{\bar{\{x\}},z})\subset Spec X$
Now that we know map on topological space for knowing the map between schemes we only have to give a map between $O_z\to O_K$ extending the morphism between $O_x\to K$. The proof in Liu's book uses the definition of properness to give such a map: in fact both ring are isomorphic:we have the map $Spec(O_z)\to Spec(O_K)$ we know that the generic point is in the image of this map and by properness the image is closed so this map is subjective on topological space and hence coming from a dominating map between local ring but both of this ring have the same field of fraction(because we know we have a map $$O_z\to K$$) and by general properties of valuation ring the map must be an isomorphism so its inverse is the extension we want. note that we never used openness of generic point in the proof.
Best Answer
I think it's a mistake. Liu's sentence
should be replaced by: