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.

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 $\tilde y$..

EDIT: This is wrong, as Mike points out, but the proof can easily be adapted to work when $X$ is Hausdorff:

For the first half above, when $X$ is Hausdorff, if $x$ and $y$ are not equal, one may choose small enough disjoint opens in $X,$ $U$ and $V$ over which there exists sections $$f \in F(U)$$ and $$g \in F(V)$$ such that $\tilde x=germ_x f$ and $\tilde y=germ_y g.$ Then $f(U)$ and $g(V)$ are necessarily disjoint and contain $\tilde x$ and $\tilde y$ respectively. If $x=y,$ then one has that there exists an open $U$ containing $x$ and $$f,g\in F(U)$$ such that $\tilde x=germ_x f$ and $\tilde y=germ_x g.$ Since $\tilde x \ne \tilde y,$ one has that $x \in \left(z \in U| germ_zf \ne germ_z g\right)=:W_x$ which is open. Hence, $f(W_x)$ and $g(W_x)$ are disjoint neighborhoods of $\tilde x$ and $\tilde y.$

(Notice that the converse still holds as stated, without assuming $X$ Hausdorff, so, it follows that the condition is still necessarily when $X$ is not Hausdorff, but not necessarily sufficient.)

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.