Yes, the geometric and the analytic versions of the Hahn–Banach theorem follow from each other. Here's a proof of the direction you ask about:
Consider the space $X = E \times \mathbb{R}$ equipped with the product topology of the topology induced by $p$ on $E$ and the ordinary topology on $\mathbb{R}$. Note that (continuous) linear functionals $g$ on $E$ bijectively correspond to (closed) linear hyperplanes of $X$ not containing $0 \times \mathbb{R}$ via the identification of $g$ with its graph $\{(x,g(x))\,:\,x \in E\}$.
Let $\Gamma = \{(x,f(x))\,:x \in F\}$ be the graph of $f$. Then $\Gamma$ is disjoint from the convex cone
$$
C = \{(x,t)\,:\,p(x) \lt t\} \subset X
$$
since $f$ is dominated by $p$: if $(x,t) \in \Gamma$ then $t = f(x) \leq p(x)$, so $(x,t) \notin C$.
Since the cone $C$ is open in $X$, the geometric version of the Hahn–Banach theorem implies that there exists a hyperplane $H \supset \Gamma$ of $X$, disjoint from $C$. In particular we must have for all $(x,t) \in H$ that $t \leq p(x)$. Note that $(0,0) \in \Gamma \subset H$, so $H$ is a linear subspace. Since $(0,1) \in C$ we have that $0 \times \mathbb{R}$ is not contained in $H$, so the hyperplane $H$ is the graph of a linear functional $g$ on $E$. Since $H \supset \Gamma$, we have found the graph of an extension $g$ of $f$ satisfying $g(x) \leq p(x)$ for all $x \in X$.
Note: It is possible to prove the geometric form of the Hahn–Banach theorem by a direct application of Zorn's lemma, see e.g. Schaefer's book on topological vector spaces, Chapter II, Theorem 3.1, page 46.
The statement you want to prove is the classic first version of the Hahn-Banach Separation Theorem .
Usually one has available the Hahn-Banach Theorem, in the form
If $M\subset X$ is a subspace, $\varphi:M\to\mathbb R$ is linear, and $\mu$ is a real seminorm on $X$ (subadditive, $\mu(\lambda x)=\lambda\,\mu(x)$ for $\lambda\geq0$) with $|\varphi(x)|\leq\mu(x)$ for all $x\in M$, then there exists $\tilde\varphi:X\to\mathbb R$, linear, with $\tilde\varphi|_M=\varphi$ and $|\tilde\varphi(x)|\leq\mu(x)$ for all $x$.
Since the interior of a convex set is convex, we may assume that $E$ is open (as the inequality will hold for any limit point). Fix $e'\in E$, $f'\in F$; The condition $E\cap F=\varnothing$ guarantees $e'\ne f'$. Then the set $Z=E-F+f'-e'$ is convex, open and $0\in E-F+f'-e'$. If $\mu$ is its Minkowski Functional, the conditions on $Z$ (open convex neighbourhood of $0$) guarantee that
$$\tag1
Z=\{x\in X:\ \mu(x)<1\}.
$$
On the subspace $M=\mathbb R(f'-e')$, define $\varphi:M\to\mathbb R$ by $\varphi(\lambda(f'-e'))=\lambda$. Then $\varphi$ is linear. As $f'-e'\not\in Z$, we have $\mu(f'-e')\geq1$. So
$$
|\varphi(\lambda(f'-e'))|=|\lambda|\leq|\lambda|\,\mu(f'-e')=\mu(\lambda(f'-e')).
$$
That is, $|\varphi(x)|\leq\mu(x)$ for all $x\in M$. By the Hahn-Banach Theorem, there exists $\tilde\varphi:X\to\mathbb R$, linear, with $|\tilde\varphi(x)|\leq\mu(x)$ for all $x\in X$.
Now, for any $e\in E$, $f\in F$,
$$
\tilde\varphi(e)-\tilde\varphi(f)+1=\tilde\varphi(e-f+f'-e)<1
$$
by $(1)$. Thus
$$\tag2
\tilde\varphi(e)<\tilde\varphi(f),\qquad e\in E,\ f\in F.
$$
Let $c=\sup\{\tilde\varphi(e):\ e\in E\}$. Then $c\leq\tilde\varphi(f)$ for all $f\in F$. Using that $F$ is a subspace, we get $c\leq t\tilde\varphi(f)$ for all $t\in\mathbb R$. This forces $\tilde\varphi(f)=0$ for all $f\in F$. Thus
$$\tag3
\tilde\varphi(f)=0,\ \tilde\varphi(e)\leq0,\qquad f\in F,\ e\in E.
$$
Finally, limit points of $E$ are also limit points of $Z$, and so they satisfy $\mu(x)\leq1$ as the Minkowski Functional is continuous. So the argument above still works, just with $\leq$ instead of $<$.
Best Answer
A complete proof (with multiple lemmas) is given in
Haïm Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, p. 4-8.