Recall that a relation is $R\subset X\times X$ and $a~b$ iff $(a,b)\in R$.
According to Charles Rezk, given a space $X$, we can construct its $k$-Hausdorffification as follows, for any relation ~$_{\alpha}$ with the property $X$/~$_{\alpha}$ is a $k$-Hausdorff space consider the quotient map $X\rightarrow X$/~$_{\alpha}$. Let $X_{\alpha}$ denote $X$/~$_{\alpha}$. Then it induces by universal property of the product a map $h:X\rightarrow \Pi_{\alpha \text{ such that }X_{\alpha}\text{ is } k\text{-Hausdorff}}X_{\alpha}$. Let $\mathbb{h}X$ be the image of $h$ and $\mathbb{h}X$ is called the $k$-Hausdorffification of $X$.
Reza also states in proposition 4.9 that if we restrict map $h$ to its image and denote such map by $q:X\rightarrow \mathbb{h}X$. Then we can construct corresponding quotient map $g:X\rightarrow X'$
First question: What is $\mathbb{h}X$? If the map to every factor is surjective, isn't the induced map by universal property of the product also surjective? So I think $\mathbb{h}X=\Pi_{\alpha \text{ such that }X_{\alpha}\text{ is } k\text{-Hausdorff}}X_{\alpha}$ but that does not seem to be true.
Second question: What is map $g$ and $X'$. I think for every surjective map $q:A\rightarrow B$ we can forget the topology on $B$ and equip $B$ with quotient topology induced by $q$ and that is also what Rezk means. Am I mistaken?
You can find in page 5 of the following link
https://ncatlab.org/nlab/files/Rezk_CompactlyGeneratedSpaces.pdf
For $(2)$, Rezk defines that $X'$ is $X/\sim$ where $x\sim y$ iff. $q(x)=q(y)$ ($q:X\to\mathrm{h}X$ being the aforeconstructed map) and defines $g:X\to X'$ to be the quotient map for this relation. I don't know what you mention $A$ and $B$ and changing the topology on $B$, $B$ ($=X'$) has defined to have the quotient topology and $g$ has been defined to be the quotient map.
For $(1)$, it is not true that for sets $X_i:i\in I$ and surjections $Y\to X_i$ that the product pairing $Y\to\prod_{i\in I}X_i$ is surjective. For a simple counterexample, let each $X_i=Y$ and let each surjection $Y\to X_i=Y$ be the identity map. If $Y$ has more than one element, the product pairing $Y\to\prod_{i\in I}X_i$ is just the diagonal... which is clearly not surjective.