$\newcommand{\con}{\operatorname{conv}}$The first calculation is correct. Firstly, some facts about $A_+$. Let $a,b\in A_+$ so $a,b\geq0$ and so $a\geq 0 \implies a+b\geq b \implies a+b\geq 0$. Also $t\in\mathbb R$ with $t\geq 0$ implies that $tA_+ \subseteq A_+$. It follows that $A_+$ is convex, and so also $-A_+$ is convex.

We are proving (i)$\implies$(ii) so by assumption $F$ is a hereditary convex closed set. By definition, $F-A_+ = \{ f-a : f\in F, a\in A_+\}$. As $F$ and $A_+$ are convex, \begin{align*} \con(F\cup(-A_+)) &= \{ tf + s(-a) : f\in F,a\in A_+, s,t\geq0, s+t=1\} \\ &= \{ tf - a : f\in F, a\in A_+, 0\leq t\leq 1\} \end{align*} as $sa\in A_+$ and $0\in A_+$. As $F$ is hereditary, given $f\in F$, as $0\leq 0\leq f$, we have $0\in F$. Then for $0\leq t\leq 1$, by convexity, $tf = (1-t)0 + tf \in F$. So $\con(F\cup(-A_+)) = F-A_+$.

For any subsets $X,Y$ we have $(X\cup Y)^\circ = X^\circ \cap Y^\circ$, just from the definition. So it remains to argue that $(-A_+)^\circ = A_+^*$. If $\omega\in (-A_+)^\circ$ then $\omega(-ta) \leq 1$ for all $t\geq0,a\in A_+$ so $\omega(a)\geq -1/t$ for all $t>0$ so $\omega(a)\geq 0$. Conversely, if $\omega(A_+)\subseteq[0,\infty)$ then $\omega(-a)\leq 0\leq 1$ for all $a\in A_+$.

I think now there is a mistake. Notice that Takesaki doesn't seem to actually use condition (i)! I would argue as follows. Hahn-Banach shows that $X^{\circ\circ}$ is the closed convex hull of $X$, for any $X$. So as $F-A_+$ is convex, $(F-A_+)^{\bar{}} = (F-A_+)^{\circ\circ}$. By the previous calculation, this equals $(F^{\wedge})^\circ$. Now use that (i) holds, so $F = (F-A_+)^{\bar{}} \cap A_+$ so $F = (F^\wedge)^\circ \cap A_+ = F^{\wedge\wedge}$, that is, (ii).

$\newcommand{\con}{\operatorname{conv}}$[Edit: This has turned out to be more involved than I thought, and my argument is now somewhat different from the book.] Firstly, some facts about $A_+$. Let $a,b\in A_+$ so $a,b\geq0$ and so $a\geq 0 \implies a+b\geq b \implies a+b\geq 0$. Also $t\in\mathbb R$ with $t\geq 0$ implies that $tA_+ \subseteq A_+$. It follows that $A_+$ is convex, and so also $-A_+$ is convex.

We are proving (i)$\implies$(ii) so by assumption $F$ is a hereditary convex closed set. Set $F_0 = \{ tf : 0\leq t<1, f\in F \}$. As $F$ is hereditary, if $F$ is non-empty there is $f\in F$ so $0\leq0\leq f\implies 0\in F$. Thus $F_0\subseteq F$, and notice that given any $f\in F_0$ there is $0<t_0<1$ and $f_0\in F_0$ with $f = t_0f_0$. $F_0$ is convex.

By definition, $F-A_+ = \{ f-a : f\in F, a\in A_+\}$. As $F_0$ and $A_+$ are convex, \begin{align*} \con(F_0\cup(-A_+)) &= \{ tf + s(-a) : f\in F_0,a\in A_+, s,t\geq0, s+t=1\} \\ &= \{ tf - a : f\in F_0, a\in A_+, 0\leq t<1 \} \cup F_0, \end{align*} as $A_+$ is invariant under scaling by a positive number. We union with $F_0$ to correspond to the case $t=1$, but as $0\in A_+$, we can write $F_0\ni f = f + 0$. Given $f-a\in F_0-A_+$, pick $t_0,f_0$ with $f = t_0 f_0$ so $f-a = t_0f_0 + (1-t_0)(-(1-t_0)^{-1}a)$. Hence $$ \con(F_0\cup(-A_+)) = F_0 - A_+. $$

For any subsets $X,Y$ we have $(X\cup Y)^\circ = X^\circ \cap Y^\circ$, just from the definition. So $$ (F_0 - A_+)^\circ = (F_0\cup(-A_+))^\circ = F_0^\circ \cap (-A_+)^\circ. $$ We first show that $(-A_+)^\circ = A_+^*$. If $\omega\in (-A_+)^\circ$ then $\omega(-ta) \leq 1$ for all $t\geq0,a\in A_+$ so $\omega(a)\geq -1/t$ for all $t>0$ so $\omega(a)\geq 0$. Conversely, if $\omega(A_+)\subseteq[0,\infty)$ then $\omega(-a)\leq 0\leq 1$ for all $a\in A_+$. Now consider \begin{align*} F_0^\circ &= \{ \omega : t\omega(f)\leq 1 \ (0\leq t<1, f\in F) \} \\ &= \{ \omega : \omega(f)\leq 1 \ (f\in F) \} = F^\circ. \end{align*} So $(F_0 - A_+)^\circ = F^\circ \cap A_+^* = F^\wedge$.

We now continue; we finally get to use condition (i)! Hahn-Banach shows that $X^{\circ\circ}$ is the closed convex hull of $X$, for any $X$ which contains $0$. So $$ (F_0 - A_+)^{\bar{}} = (F_0 - A_+)^{\circ\circ} = (F^\wedge)^\circ. $$ As $F_0 \subseteq F$ and is dense, we have $(F - A_+)^{\bar{}} \subseteq (F_0 - A_+)^{\bar{}} \subseteq (F - A_+)^{\bar{}}$ and so we have equality. Thus $(F - A_+)^{\bar{}} = (F^\wedge)^\circ$. Now use that (i) holds, so $F = (F-A_+)^{\bar{}} \cap A_+$ so $F = (F^\wedge)^\circ \cap A_+ = F^{\wedge\wedge}$, that is, (ii).

$\newcommand{\con}{\operatorname{conv}}$The first calculation is correct. Firstly, some facts about $A_+$. Let $a,b\in A_+$ so $a,b\geq0$ and so $a\geq 0 \implies a+b\geq b \implies a+b\geq 0$. Also $t\in\mathbb R$ with $t\geq 0$ implies that $tA_+ \subseteq A_+$. It follows that $A_+$ is convex, and so also $-A_+$ is convex.

We are proving (i)$\implies$(ii) so by assumption $F$ is a hereditary convex closed set. By definition, $F-A_+ = \{ f-a : f\in F, a\in A_+\}$. As $F$ and $A_+$ are convex, \begin{align*} \con(F\cup(-A_+)) &= \{ tf + s(-a) : f\in F,a\in A_+, s,t\geq0, s+t=1\} \\ &= \{ tf - a : f\in F, a\in A_+, 0\leq t\leq 1\} \end{align*} as $sa\in A_+$ and $0\in A_+$. As $F$ is hereditary, given $f\in F$, as $0\leq 0\leq f$, we have $0\in F$. Then for $0\leq t\leq 1$, by convexity, $tf = (1-t)0 + tf \in F$. So $\con(F\cup(-A_+)) = F-A_+$.

For any subsets $X,Y$ we have $(X\cup Y)^\circ = X^\circ \cap Y^\circ$, just from the definition. So it remains to argue that $(-A_+)^\circ = A_+^*$. If $\omega\in (-A_+)^\circ$ then $\omega(-ta) \leq 1$ for all $t\geq0,a\in A_+$ so $\omega(a)\geq -1/t$ for all $t>0$ so $\omega(a)\geq 0$. Conversely, if $\omega(A_+)\subseteq[0,\infty)$ then $\omega(-a)\leq 0\leq 1$ for all $a\in A_+$.

I guess the other claims follow similarly; I'll try to think later.

$\newcommand{\con}{\operatorname{conv}}$The first calculation is correct. Firstly, some facts about $A_+$. Let $a,b\in A_+$ so $a,b\geq0$ and so $a\geq 0 \implies a+b\geq b \implies a+b\geq 0$. Also $t\in\mathbb R$ with $t\geq 0$ implies that $tA_+ \subseteq A_+$. It follows that $A_+$ is convex, and so also $-A_+$ is convex.

We are proving (i)$\implies$(ii) so by assumption $F$ is a hereditary convex closed set. By definition, $F-A_+ = \{ f-a : f\in F, a\in A_+\}$. As $F$ and $A_+$ are convex, \begin{align*} \con(F\cup(-A_+)) &= \{ tf + s(-a) : f\in F,a\in A_+, s,t\geq0, s+t=1\} \\ &= \{ tf - a : f\in F, a\in A_+, 0\leq t\leq 1\} \end{align*} as $sa\in A_+$ and $0\in A_+$. As $F$ is hereditary, given $f\in F$, as $0\leq 0\leq f$, we have $0\in F$. Then for $0\leq t\leq 1$, by convexity, $tf = (1-t)0 + tf \in F$. So $\con(F\cup(-A_+)) = F-A_+$.

For any subsets $X,Y$ we have $(X\cup Y)^\circ = X^\circ \cap Y^\circ$, just from the definition. So it remains to argue that $(-A_+)^\circ = A_+^*$. If $\omega\in (-A_+)^\circ$ then $\omega(-ta) \leq 1$ for all $t\geq0,a\in A_+$ so $\omega(a)\geq -1/t$ for all $t>0$ so $\omega(a)\geq 0$. Conversely, if $\omega(A_+)\subseteq[0,\infty)$ then $\omega(-a)\leq 0\leq 1$ for all $a\in A_+$.

I think now there is a mistake. Notice that Takesaki doesn't seem to actually use condition (i)! I would argue as follows. Hahn-Banach shows that $X^{\circ\circ}$ is the closed convex hull of $X$, for any $X$. So as $F-A_+$ is convex, $(F-A_+)^{\bar{}} = (F-A_+)^{\circ\circ}$. By the previous calculation, this equals $(F^{\wedge})^\circ$. Now use that (i) holds, so $F = (F-A_+)^{\bar{}} \cap A_+$ so $F = (F^\wedge)^\circ \cap A_+ = F^{\wedge\wedge}$, that is, (ii).