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
$$\psi(x,y) = \left\langle (x,y), \left(\frac15, \frac25\right)\right\rangle = \frac{x}2 + \frac{2y}5, \quad\forall (x,y)\in\mathbb{R}^2$$
I'll take real spaces for simplicity.
As for why local convexity is needed in the corollaries: The proof of Corollary 1 states "In virtue of the Hahn-Banach theorem[...]". But notice that geometric HB deals with an open convex set, and analytical HB deals with a seminorm, none of which we have during the proof! So what's the deal? Some details are being ommited. When we expand them, the necessity for local convexity becomes clear:
Let us try to use geometric HB: By local convexity, and since $x_0$ does not belong to the closure $\overline{M_0}$, there exists a convex open set containing $x_0$ and which does not intersect $M_0$. Then we can use geometric HB directly, separating $\overline{M_0}$ and $x_0$ by a closed hyperplane $N$. So the quotient $M/N$ is a Hausdorff one-dimensional TVS, hence topologically isomorphic to $\mathbb{R}$ (Treves, Theorem 9.1(a)). Let $g\colon M/N\to\mathbb{R}$ be a linear topological isomorphism taking $x_0+N$ to $1$. Then the composition of $g$ with the canonical quotient map $M\to M/N$ is nonzero (because it takes $x_0$ to $1$) but vanishes on $M_0$.
We can also use analytical HB: Since $M$ is locally convex, $M/\overline{M_0}$ is also locally convex. Denote $\phi\colon M\to M/\overline{M_0}$ the quotient map. Since $\phi(x_0)\neq 0$, $M/\overline{M_0}$ is Hausdorff and locally convex, there exists a continuous seminorm $p$ on $M/\overline{M_0}$ such that $p(\phi(x_0))\neq 0$. Then proceed in the same manner as Treves' original proof.
As for why the Hausdorff property implies that $f$ is continuous, there are two options:
You have probably seen that every finite-dimensional vector space admits a unique Hausdorff topological vector space topology. So the topology of $\mathbb{R}\phi(x_0)$ is the one coming from the linear isomorphism $f$, which automatically makes it continuous.
Alternatively, since we're dealing with locally convex spaces, we can take a continuous seminorm $p$ on $M/\overline{M_0}$ for which $p(\phi(x_0))=1$. Thus $p(\lambda\phi(x_0))=|\lambda|$, from which continuity of $f'$ follows.
To see why we need the Hausdorff property in Corollary $2$, we should try to mimic the proof of Corollary 1: If $x_0\neq 0$, define $f'\colon\mathbb{R}x_0\to\mathbb{R}$ as $\lambda x_0\mapsto x_0$, and extend it by HB to all of $M$. But the problem is that the map $f'$ is the application suppose that $V$ is a non-Hausdorff topological vector space. This means that $\left\{0\right\}$ is not a closed set. Let $y_0$ be any nonzero vector in $\overline{\left\{0\right\}}$. Since $\overline{\left\{0\right\}}$ is the closure of a subspace, the it is a subspace as well, so the line $\mathbb{R}y_0$ passing through $y_0$ is contained in it. Moreover, the zero vector $0$ is dense in $\mathbb{R}y_0$, from which you can conclude that the subspace topology of $\mathbb{R}y_0$ is the undiscrete one: Only $\varnothing$ and $\mathbb{R}y_0$ are open. Therefore the map $\lambda y_0\mapsto y_0$ is not continuous on $\mathbb{R}y_0$.
As for the last question, I do not know any direct way to see why the separation of points property implies the usual Hahn-Banach theorems. I tried using Minkowski functionals, but to no avail.
However, I will assume all proof of Schechter's book are correct, although I did not check them. He does indeed provide the equivalence of 26 forms of Hahn-Banach (all of which are, thus, equivalent weaker forms of the Axiom of Choice). They are numbered $(HB1)-(HB26)$. The analytic HB is (HB2); geometric HB is (HB18); Corollary 2 is (HB22)
I will not write down the statements, but the way their equivalence is proven is:
- p. 322-323:
$$(HB1)\to(HB2)\to(HB3)\to(HB1)$$
- p. 322:
$$(HB2)\to(HB4)\to(HB5)\to(HB1)$$
- p. 618-619:
$$(HB2)\to(HB7)\to(HB8)\to(HB9)\to(HB11)\to(HB12)\to(HB1)$$
- p.618-619:
$$(HB8)\to(HB10)\to(BH11)$$
- p. 620-621:
$$(HB12)\to(HB13)\to(HB14)\to(HB12)$$
- p. 717:
$$(HB15)\leftrightarrow(HB2)\leftrightarrow(HB16)$$
- p. 754:
$$(HB4)\to(HB17)\to(HB18)\to(HB19)\to(HB20)\to(HB21)\to(HB22)\to(HB9)$$
- p. 754:
$$(HB20)\to(HB23)\to(HB11)$$
- p. 760:
$$(HB20)\to(HB24)\to(HB21)$$
- p. 805-806:
$$(HB2)\to(HB25)\to(HB26)\to(HB12)$$
So the shortest path from (HB22) to either (HB2) or (HB18) is
$$(HB22)\to(HB9)\to(HB11)\to(HB12)\to(HB1)\to(HB2)$$
The first implications are basically trivial, however (HB12) deals with "Luxembourg's measure" and (HB1) with Banach limits, so there is no easy way to adapt his proof.
Best Answer
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.