Corollary III.9.4 in Hartshorne's Algebraic Geometry: Let $f: X \to Y$ be a separated morphism of finite type of noetherian schemes with $Y$ affine, and let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. For any point $y \in Y$, let $X_y$ denote the fiber over $y$, and let $\mathcal{F}_y$ be the induced sheaf. On the other hand, let $k(y)$ denote the constant sheaf $k(y)$ on the closed subset $\{y \}^-$ of $Y$. Then for all $i \ge 0$, there exists a natural isomorphism $$H^i(X_y, \mathcal{F}_y) \cong H^i(X, \mathcal{F} \otimes k(y)) $$
$\textbf{Proof}:$ First let $Y' \subseteq Y$ be the reduced induced subscheme structure on $\{y\}^-$, and let $X'= X \times_Y Y'$, which is a closed subscheme of $X$. The next line of the proof I have been struggling to understand for a while now, "then both sides of our desired isomorphism depend only on the sheaf $\mathcal{F}'= \mathcal{F} \otimes k(y)$ on $X'$. Thus we can replace $X,Y, \mathcal{F}$ by $X', Y', \mathcal{F'}$..".
Where is this sheaf $\mathcal{F}'$ appearing from??
Moreover, it seems when we have a scheme of the form $X \times Y \to Y$ it is ok (see Exercise 12.6) to describe a fiber as $X \times \{y \}$ as opposed to $X \times_Y \operatorname{Spec} k(y)$. Confusingly, in this corollary he actually choses to replace the entire family $X \to Y$ by $X \times_Y \{y \} \to \{y\}$.