Let $X,Y$ be schemes. Let $f:X\rightarrow Y$, $g:Y\rightarrow Y$ be morphisms of schemes. Then we can form the base change

\begin{CD} X\times_{f,Y,g} Y @>>> Y\\ @VVV @VV{g}V \\X @>{f}>> Y \end{CD}

Therefore, for any morphism $U\rightarrow X$, we can form the fibre product $U\times_{Y,g} Y$ as follows:

\begin{CD}U\times_{Y,g}Y @>>> X\times_{f,Y,g} Y @>>> Y\\ @VVV @VVV @VV{g}V \\ U @>>> X @>{f}>> Y \end{CD}

Let $X\rightarrow X$ be the identity morphism. Then we have an induced map $X\rightarrow X\times_{f,Y,g} Y$ by the universal property of $X\times_{f,Y,g} Y$. Then the base change

\begin{CD} U\times_{X,f} X @>>> X\\ @VVV @VVV \\ U\times_{Y,g}Y @>>> X\times_{f,Y,g} Y \end{CD}

gives us the desired base change map $U\times_{X}X\rightarrow U\times_{Y}Y$.