The answer of the first question is that for all $U$ open in $X$, you need that for all $f,g \in F(U)$ the subset $\left(x \in U| germ_xf \ne germ_xg\right) \subset U$ is open.
If this holds, then for any two points $\tilde x$ and $\tilde y$ in $E(F),$ one may take two opens $U$ of $x$ and $V$ of $y$ respectively, where $x$ and $y$ are their images in $X,$ such that there exists $f \in F(U)$ and $g \in F(V)$ such that $$germ_xf=\tilde x$$ and $$germ_yg=\tilde y.$$ Then the subset $W$ of $U \cap V$ on which $f|_{U\cap V}$ and $g|_{U \cap V}$ agree is closed, and then one may define the open sets $\tilde U:=U - W$ and $\tilde V:V -W.$ One then has that $f(\tilde U)$ and $g(\tilde V)$ are disjoint opens of $\tilde x$ and $\tilde y$.
Conversely, suppose that $E(F)$ is Hausdorff, and let $U$ be an open of $X.$ Consider for all $f,g \in F(U)$ the subset $Q:=\left(x \in U| germ_xf \ne germ_xg\right) \subset U$. Let $$z_1=germ_x f \ne z_2=germ_x g.$$ There exists disjoint neighborhoods $V_1$ and $V_2$ respectively. Let $$O_x:=f^{-1}(V_1) \cap g^{-1}(V_2).$$ This is a neighborhood of $x.$ $f(O_x) \subset V_1$ and $g(O_x) \subset V_2$ are now also disjoint, hence $O_x \subset Q,$ so $Q$ is open.