I'm reading the proof of the following statement in Milne's book "Algebraic geometry".
Chow's Lemma: Let $V$ be a complete irreducible variety. There exists a projective variety $V'$ and a surjective map $f:V'\to V$ such that for some dense open subset $U\subseteq V$ the map $f:f^{-1}(U)\to U$ is an isomorphism.
At page 166 of the book the author says (I use an image because MSE doesn't support tikzcd diagrams)
First of all I don't understand what is the upper map of the first diagram. Secondly, I'm really not familiar with "cartesian diagrams" and "pullbacks/pushouts". Could you explicitate this reasoning a bit more?
(The author is trying to prove that $f^{-1}(U_i)\subseteq q^{-1}(V_i)$.)
I must confess that I have not gone through the proof that you recommended I go through, but from what I can see on screen, I can try and intuitively explain what fiber product does. In simple terms, as the name suggests, it is trying to tell us the fiber (i.e. inverse image) of a morphism. There are other uses of fiber product like base change, but since I am going to skip the technicalities, I'll ignore that too.
Let us fix a morphism $f:X\rightarrow Y$, which we shall treat as our "primary" morphism. Then, if $i:Z\hookrightarrow Y$ is some subscheme, then the fiber product which is denoted by $X\times_Y Z$ and accompanied buy a commutative diagram $$\require{AMScd}\begin{CD} X\times_Y Z @> >> Z\\ @VVV @VViV \\ X @>f>> Y \end{CD}$$ as well as some universal property, is the unique subscheme of $X$ whose image under the map $f$ is contained in $Z$, and is maximal with this property. In other words, $X\times_Y Z$ can be sort of thought as $f^{-1}(Z)$. As one would expect, if $Z$ is a closed (resp. open) subsheme of $Y$, then the fiber product is also a closed (resp. open) subscheme of $X$. In fact, if $Z$ is an open subscheme, then $X\times_Y Z$ is exactly the open subscheme $f^{-1}(Z)\subset X$.
Rather than taking an open/closed subscheme, another important case, is when we assume $Z=$Spec$(\kappa (y))$ for some $y\in Y$. Then, the fiber product gives a natural scheme structure on $f^{-1}(y)$.
Now, a diagram $$\require{AMScd}\begin{CD} W @> >> Z\\ @VVV @VViV \\ X @>f>> Y \end{CD}$$ is said to be cartesian if there is an isomorphism $W\cong X\times_Y Z$ in such a way that the two commutative squares are "exactly the same".
I am not sure if this is the answer you are looking for, but I hope this gives you some tools to understand the proof.