Proof.
Part (1) follows from the injectivity of the base change map discussed in Section 62.3. (This argument works as long as $S' \to S$ is surjective.)
Let $\alpha '$ be as in (2). Denote $\alpha '' = \text{pr}_1^*\alpha ' = \text{pr}_2^*\alpha '$ the common value.
Let $(X/S)^{(r)}$ be the set of $x \in X$ with $\text{trdeg}_{\kappa (f(x))}(\kappa (x)) = r$ and similarly define $(X'/S')^{(r)}$ and $(X''/S'')^{(r)}$ Taking coefficients, we may think of $\alpha '$ and $\alpha ''$ as functions $\alpha ' : (X'/S')^{(r)} \to \mathbf{Z}$ and $\alpha '' : (X''/S'')^{(r)} \to \mathbf{Z}$. Given a function
\[ \varphi : (X/S)^{(r)} \to \mathbf{Z} \]
we define $g^*\varphi : (X'/S')^{(r)} \to \mathbf{Z}$ by analogy with our base change operation. Namely, say $x' \in (X'/S')^{(r)}$ maps to $x \in X$, $s' \in S'$, and $s \in Z$. Denote $Z' \subset X'_{s'}$ and $Z \subset X_ s$ the integral closed subschemes with generic points $x'$ and $x$. Note that $\dim (Z') = r$. If $\dim (Z) < r$, then we set $(g^*\varphi )(x') = 0$. If $\dim (Z) = r$, then $Z'$ is an irreducible component of $Z_{s'}$ and hence has a multiplicity $m_{Z', Z_{s'}}$. Call this $m(x', g)$. Then we define
\[ (g^*\varphi )(x') = m(x', g) \varphi (x) \]
Note that the coefficients $m(x', g)$ are always positive integers (see for example Lemma 62.3.1). We similarly have base change maps
\[ \text{pr}_1^*, \text{pr}_2^* : \text{Map}((X'/S')^{(r)}, \mathbf{Z}) \longrightarrow \text{Map}((X''/S'')^{(r)}, \mathbf{Z}) \]
It follows from the associativity of base change that we have $\text{pr}_1^* \circ g^* = \text{pr}_2^* \circ g^*$ (small detail omitted). To be explicity, in terms of the maps of sets this equality just means that for $x'' \in (X''/S'')^{(r)}$ we have
\[ m(x'', \text{pr}_1) m(\text{pr}_1(x''), g) = m(x'', \text{pr}_2) m(\text{pr}_2(x''), g) \]
provided that $\text{pr}_1(x'')$ and $\text{pr}_2(x'')$ are in $(X''/S'')^{(r)}$. By Lemma 62.5.8 and an elementary argument1 using the previous displayed equation, it follows that there exists a unique map
\[ \alpha : (X/S)^{(r)} \to \mathbf{Q} \]
such that $g^*\alpha = \alpha '$. To finish the proof it suffices to show that $\alpha $ has integer values (small detail omitted: one needs to see that $\alpha $ determines a locally finite sum on each fibre which follows from the corresponding fact for $\alpha '$). Given any $x \in (X/S)^{(r)}$ with image $s \in S$ we can pick a point $s' \in S'$ such that $\kappa (s')/\kappa (s)$ is separable. Then we may choose $x' \in (X'/S')^{(r)}$ mapping to $s$ and $x$ and we see that $m(x', g) = 1$ because $Z_{s'}$ is reduced in this case. Whence $\alpha (x) = \alpha '(x')$ is an integer.
$\square$
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