A counterexample.
It is not true that a poset, in which every chain in order-separable, necessarily admits a strictly increasing real-valued function. In fact, here is an example of a poset $(X,\preceq)$ of cardinality $\aleph_1$ which does not admit a strictly increasing real-valued function, even though every chain in $(X,\preceq)$ is finite.
Let $X=\omega_1\times\omega_1$. For $(\alpha,\beta),(\alpha',\beta')\in X$, define
$$(\alpha,\beta)\prec(\alpha',\beta')\iff\alpha\lt\alpha'\text{ and }\beta\gt\beta'$$
and
$$(\alpha,\beta)\preceq(\alpha',\beta')\iff(\alpha,\beta)\prec(\alpha',\beta')\text{ or }(\alpha,\beta)=(\alpha',\beta').$$
Since $(\omega_1,\lt)$ is well-ordered, it's clear that $(X,\preceq)$ contains no infinite chain. Assume for a contradiction that there is strictly increasing function $f:X\to\mathbb R$.
For each $\alpha\in\omega_1$ there exists $\beta_\alpha\in\omega_1$ such that $\{\xi\in\omega_1:f(\alpha,\xi)\ge f(\alpha,\beta_\alpha)\}$ is uncountable. Let $t_\alpha=f(\alpha,\beta_\alpha)$. Choose $\alpha,\alpha'\in\omega_1$ so that $\alpha\lt\alpha'$ and $t_\alpha\ge t_{\alpha'}$. Let $\beta'=\beta_{\alpha'}$, so that $f(\alpha',\beta')=t_{\alpha'}$. Choose $\beta\in\omega_1$ so that $\beta\gt\beta'$ and $f(\alpha,\beta)\ge f(\alpha,\beta_\alpha)=t_\alpha$. Then $(\alpha,\beta)\prec(\alpha',\beta')$ and $f(\alpha,\beta)\ge t_\alpha\ge t_{\alpha'}=f(\alpha',\beta')$, contradicting our assumption that $f$ is strictly increasing.
Remark. More generally, for any chain $Y$, a strictly increasing function $f:X\to Y$ exists if and only if $Y$ contains a subset of order type $\omega_1$ or $\omega_1^*$.
A characterization.
An order-ideal in a poset $(X,\preceq)$ is a set $A\subseteq X$ which is closed downward, i.e., if $x\in X$ and $x\preceq a\in A$, then $x\in A$.
A poset $(X,\preceq)$ admits a strictly increasing real-valued function if and only if there is a countable family $(A_n:n\in\mathbb N)$ of order-ideals such that, whenever $x_1,x_2\in X$ and $x_1\prec x_2$, there is some $n\in\mathbb N$ such that $x_1\in A_n$ and $x_2\notin A_n$.
Given such order-ideals, we can define a strictly increasing real-valued function by setting
$$f(x)=\sum_{x\notin A_n}2^{-n}.$$
Conversely, given a strictly increasing function $g:X\to\mathbb R$, we can take the order-ideals $A_n=\{x\in X:g(x)\lt r_n\}$ where $\mathbb Q=\{r_n:n\in\mathbb N\}$ is some enumeration of the rational numbers.
There is nothing unusual about the Boolean algebra $A$; it is just the usual power set Boolean algebra of the set $2^N$, with union and intersection and set complement as its operations. The fact that $2^N$ happens to itself be a power set is irrelevant.
It appears that there is either a typo or a rather unfortunate abuse of notation that is confusing you: when it refers to $S$ as being an atom of $A$, it should indeed be $\{S\}$ instead. Similarly, the definition of $A_i$ should be $$A_i = \bigcup_{\{i\}\subseteq S \subseteq N}\{S\}.$$ In other words, $A_i$ is just the set of all subsets of $N$ that contain $i$.
So, this means that for $S\subseteq N$, $\bigcap_{i\in S}A_i$ is just the set of all $T\subseteq N$ such that $S\subseteq T$. We can write this as the union of the singletons $\{T\}$ for each such $T$. So, since $p$ is additive, $$p\left(\bigcap_{i\in S}A_S\right)=\sum_{T\supseteq S}p(\{T\})$$ and thus $$p_\epsilon(A_1,\dots,A_n)=\sum_{S\subseteq N}\epsilon(S)\sum_{T\supseteq S}p(\{T\})=\sum_{T\subseteq N}\sum_{S\subseteq T}\epsilon(S)p(\{T\}).$$
Best Answer
You are right in both points.
A possible way to define your desired map is to take $$S \mapsto (u_1, \ldots, u_n),$$ where $u_i \in \{0,1\}$ and $u_i = 1$ iff $i \in S$.