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.
A simple corollary which I studied, for example,is that for every $x\in V$ you have that $\left\lVert x\right\rVert_V = \sup_{f\in B_{V'}} |f(x)|=\max_{f\in B_{V'}} |f(x)|$, which is nice since in general the "$\sup$" is not achieved.
Anyway I'll let more experienced mathematicians answering you better, for instance you could try to prove this ;)
Best Answer
In case of $(\mathbb{R}^2, \|\cdot\|_2)$, things are quite simple.
Let $U \le \mathbb{R}^2$ be a subspace and $\phi : U \to \mathbb{R}$ a linear functional. By the Riesz representation theorem, there exists $a \in U$ such that $\phi(x) = \langle x, a\rangle, \forall x \in U$. Then the linear functional $\psi : \mathbb{R}^2 \to \mathbb{R}$ given by the same formula $\psi(x) = \langle x, a\rangle, \forall x \in \mathbb{R}^2$ is the unique Hahn-Banach extension of $\phi$.
Namely, clearly $\psi$ extends $\phi$ and $\|\phi\| = \|a\|_2 = \|\psi\|$ so $\psi$ is a Hahn-Banach extension of $\phi$.
Let $\zeta : \mathbb{R}^2 \to \mathbb{R}$ be another Hahn-Banach extension of $\phi$. By the Riesz representation theorem, there exists $b \in \mathbb{R}^2$ such that $\zeta(x) = \langle x, b\rangle,\forall x \in \mathbb{R}^2$. Since $\zeta$ extends $\phi$, we have
$$\langle x, a\rangle = \phi(x) = \zeta(x) = \langle x, b\rangle, \forall x \in U \implies \langle x, a - b\rangle = 0, \forall x \in U \implies a - b \perp U$$
Since $b = \underbrace{a}_{\in U} + \underbrace{(b - a)}_{\in U^\perp}$, the Pythagorean theorem gives
$$\|a\|_2^2 + \|b - a\|_2^2 = \|b\|_2^2 = \|\zeta\|^2 = \|\phi\|^2 = \|a\|_2^2 \implies b - a = 0 \implies a = b$$
Therefore $\zeta = \phi$.
This explicit construction is in fact always possible when dealing with a Hilbert space, which $(\mathbb{R}^2, \|\cdot\|_2)$ is an example of.
For an explicit example of the above discussion, consider the subspace $Y = \{(x,2x) \in \mathbb{R}^2 : x \in \mathbb{R}\} \le \mathbb{R}^2$ and the linear functional $\phi :Y \to \mathbb{R}$ given by $\phi(x,y) = x$.
An orthonormal basis for $Y$ is $\left\{\frac1{\sqrt{5}}(1,2)\right\}$ so the orthogonal projection $P_Y$ onto $Y$ is given by
$$P_Y(x,y) = \left\langle (x,y),\frac1{\sqrt{5}}(1,2)\right\rangle \frac1{\sqrt{5}}(1,2) = \left(\frac{x+2y}5, \frac{2x+4y}5\right)$$
Now notice that for all $(x,y) \in Y$ we have
$$\phi(x,y) = x = \langle (x,y), (1,0)\rangle = \langle (x,y), P_Y(1,0)\rangle = \left\langle (x,y), \left(\frac15, \frac25\right)\right\rangle$$
Now the above discussion implies that the unique Hahn-Banach extension of $\phi$ is given by $\psi : \mathbb{R}^2 \to \mathbb{R}$ defined as
$$\psi(x,y) = \left\langle (x,y), \left(\frac15, \frac25\right)\right\rangle = \frac{x}2 + \frac{2y}5, \quad\forall (x,y)\in\mathbb{R}^2$$