First, you can see the conclusion in the case $d=0$ as a consequence of the fact that the property of being 'flat of relative dimension $n$' is preserved under base extension, by looking at the case where $X$ and $Y$ are varieties. One reduces to this case by the below, but base changing by the inclusion of the variety, not its generic point, into $X$.

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To perform Fulton's 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.


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How to use the algebraic lemma to conclude when $d>0$.

Let me rename the Artinian ring $A$ as in $Y'=\operatorname{Spec} A$ by $A'$. So now $Y'=\operatorname{Spec} A'$.



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 local 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 $d>0$.