How is the "canonical" homomorphism $(R^pf_*\mathcal{F})_y\otimes_{O_{Y,y}}k(y)\rightarrow H^p(X_y,\mathcal{F}_y)$ defined? Let $f:X\rightarrow Y$ be a proper morphism of Noetherian schemes, $\mathcal{F}$ a coherent $O_X$-module and $y\in Y.$ Denote by $k(y)$ the residue field of $Y$ at $y$, $X_y=X\times_Y\operatorname{Spec} k(y)$ and $\mathcal{F}_y=p^*\mathcal{F}$ where $p$ is the projection $X\times_Y\operatorname{Spec}k(y)\rightarrow X$.
How is the "canonical" homomorphism $(R^if_*\mathcal{F})_y\otimes_{O_{Y,y}}k(y)\rightarrow H^i(X_y,\mathcal{F}_y)$ defined ?
 A: This is an application of a more general fact. Let $X,Y,Y'$ be noetherian schemes, let $f:X\to Y$ be of finite type and separated, let $u:Y'\to Y$ be arbitrary, and let $X'=X\times_YY'$ be the base change, organized in the following commutative diagram:
$$\require{AMScd}
\begin{CD}
X' @>{v}>> X\\
@VV{g}V @VV{f}V \\
Y' @>{u}>> Y
\end{CD}$$
Let $\def\cF{\mathcal{F}}\cF$ be a quasi-coherent sheaf on $X$. I claim that there is a natural map $$u^*R^if_*(\cF)\to R^ig_*(v^*\cF).$$
Proof: The question is local on $Y$ and $Y'$, so we may assume that $Y=\operatorname{Spec} A$ and $Y'=\operatorname{Spec} A'$. So it suffices to demonstrate a map $H^i(X,\cF)\otimes_A A'\to H^i(X',\cF')$ where $\cF'=v^*\cF$. But as $X$ is separated and $\cF$ is quasi-coherent, we may use the Cech cohomology to compute $H^i(X,\cF)$. So let $\mathfrak{U}$ be a cover of $X$ by open affines $U_i$ and let $C(\mathfrak{U},\cF)$ be the Cech complex. Since affine morphisms are stable under base change, $v$ is also affine, and the Cech complex associated to the open cover $v^{-1}(U_i)$ and the sheaf $\cF'$ is exactly $C(\mathfrak{U},\cF)\otimes_A A'$, which computes $H^i(X',\cF')$ for the same reasons as before. This construction gives us our natural map: it's the induced map on cohomology from the chain map $C(\mathfrak{U},\cF)\to C(\mathfrak{U},\cF)\otimes_A A'$ sending $c\mapsto c\otimes 1$. $\blacksquare$
To apply this to your specific case, let $Y'=\operatorname{Spec} k(y)$ and $X'=X_y$ so the diagram is a base-change diagram along the natural inclusion $y\to Y$. Since your canonical homomorphism can be computed Zariski-locally in a neighborhood of $y$, we may assume $Y$ is affine so that $R^if_*(\cF)$ is the sheaf associated to $H^i(X,\cF)$. Then pullback along $\operatorname{Spec} k(y)\to Y$ is exactly applying $-\otimes_{\mathcal{O}_{Y,y}} k(y)$, so we've identified the LHS of your formula with the more general fact. The RHS is essentially the same: since $\operatorname{Spec} k(y)$ is affine, $R^ig_*(-)=H^i(X_y,-)$.
