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.
I think this statement is false. If there is an extension there are infinitely many whether or not $A$ is bounded. (As stated, there is no requirement that the extended linear map is bounded, even in the case $A$ is bounded).
There is a linear functional $f$ on $\overline V$ which is $0$ in $V$ but not on $\overline V$. Fix $y \neq 0$ in $Y$ and consider $B(.)+nf(.)y$ where $B$ is any extension of $A$ to $\overline V$. These are all distinct extension of $A$ to linear maps on $\overline V$.
Best Answer
Consider the case $F=M$ and $T$ the identity map on $M$. An extension of this to $T': E \to M$ is a projection of $E$ on $M$. Then with $N = \ker(T')$ you can write $E = M \oplus N$.
According to Asaf Karagila's answer here, the Axiom of Choice is equivalent to the statement that this can always be done.