To perform the reduction:
We may assume $\alpha\in Z_\ast X$ is the class of a variety $[V]$, since $f^\ast, f_\ast$ are linear. Then $f(V)=W$ is a variety. Then to compare the multiplicities of each irreducible component of $f'_\ast g'^\ast\alpha$ and $g^\ast f_\ast\alpha$ it suffices to do after looking over the generic points of $V$ and $W$, since because $g$ is flat, each irreducible component of $g^{-1}(W)$ maps to the generic point of $W$. so we may assume $X=\operatorname{Spec} L$, $Y=\operatorname{Spec} K$, in which case an irreducible component of $Y'$ becomes the spectrum of a local Artinian ring. As mentioned in the text, properness and flatness are preserved under base change.
How to use the algebraic lemma to conclude:
Let me rename the Artinian ring $A$ as in $Y'=\operatorname{Spec} A$ by $A'$. So now $Y'=\operatorname{Spec} A'$.
The algebraic lemma has rings called $A$ and $B$. First with $A=K$, $B=L$, and $M=A'\otimes_K L$, if $[L:K]=\infty$ then $A'\otimes_K L$ does not have finite length over $K$, so now replacing $A=K$, $B=A'$, and keeping $M$ the same, since $g$ is of finite type, $K\subset k(Y')$ is finite, so the algebraic lemma shows us that $A'\otimes_K L$ does not have finite length over $A'$; this shows that if $f_\ast[X]=0$, $f'_\ast[X']=0$.
Now assume $d>0$. Now $l_K(A'\otimes_K L)<\infty$ and we compute this number using the algebraic lemma applied both to $K\subset L$ and $K\subset A$ as before. It shows us that $$d\cdot l_L(A'\otimes_K L)=[k(A'):K]\cdot l_{A'}(A'\otimes_K L).$$
Now $l_L(A'\otimes_K L)=\dim_L(A'\otimes_K L)=\dim_K(A')=l_{A'}(A')\cdot[k(A'):K]$, so we find $$d\cdot l_{A'}(A')=l_{A'}(A'\otimes_K L).$$ Now if $Y'=\{P\}$ as sets, $[Y']=l_{A'}(A')\cdot P$ while $f'_\ast[X']=l_{A'}(A'\otimes_K L)\cdot P$. To see this last point, recall that since $A'\otimes_K L$ is finite over $L$, it is a direct product of finitely many Artinian rings; now apply the algebraic lemma one last time to each point in $X'$. This proves the equality $$f'_\ast[X']=d[Y']$$ in the case $d>0$.