Concerning question 2, the theorem that you mention in the case of
$\omega_1$ is not actually difficult. I had briefly sketched a
proof of it at the conclusion of my answer to the math.SE question
Linearly ordered sets somewhat
similar to $\mathbb{Q}$, which is concerned with these types of orders---what I called $\mathcal{Q}_A$ is the same as your $\Phi(A)$. Here is a complete argument:

**Theorem.** Every $\omega_1$-like dense linear ordering is
isomorphic to $\Phi(A)$ for some $A\subset\omega_1$.

Proof. Suppose that $L$ is an $\omega_1$-like dense linear order.
Let $\langle x_\alpha\mid
\alpha\lt\omega_1\rangle$ be any increasing cofinal $\omega_1$-sequence in
$L$, containing none of its limit points (i.e., scattered). Let
$\tau_\alpha$ be the interval of points above
$\cup_{\beta\lt\alpha}\tau_\beta$ and below the point $x_\alpha$. This
is either $\eta$ or $1+\eta$, depending on whether
$\cup_{\beta\lt\alpha}\tau_\beta$ has a supremum in $L$ or not. These
intervals therefore realize $L$ as
$\Sigma_{\alpha\lt\omega_1}\tau_\alpha$, as desired. Thus, $L$ is
$\Phi(A)$, where $A$ is the set of $\alpha$ where that supremum
exists. QED

In that previous answer, I proved that the $\Phi(A)$ orders are
determined up to isomorphism essentially by the equivalence of $A$
modulo the club filter.

**Theorem.** $\Phi(A)$ is isomorphic to $\Phi(B)$ if and only
if $A$ and $B$ agree on having $0$ and also agree modulo the club
filter, meaning that there is a closed unbounded set
$C\subset\omega_1$ such that $A\cap C=B\cap C$. In other words,
this is if and only if $A$ and $B$ agree on $0$ and are equivalent
to $B$ in $P(\omega_1)/\text{NS}$, as subsets modulo the
nonstationary ideal.

It would seem to be an interesting question to inquire in your
style whether one may extend this beyond $\omega_1$ to higher
cardinals.

The anwer, unfortunately, is negative. Suppose $\kappa\gt\omega_1$ is any cardinal. Let $A$ be the empty set, and let $B$ be any nonstationary set containing some ordinals with uncountable cofinality; for example, consider the singleton set $B=\{\ \omega_1\ \}$. These two sets agree on a club, since both omit a club. But meanwhile, $\Phi(A)$ and $\Phi(B)$ are not
isomorphic, because the former has all points having cofinality
$\omega$, but the latter has points of uncountable
cofinality.

All $\aleph_1$-dense sets of reals can be isomorphic, Fund. Math. 79 (1973), 101-106) and Moore (A five element basis for the uncountable linear orders, Ann. of Math. 163 (2006), 669-688). See also Todorcevic's surveyTrees and linearly ordered setsin the Handbook of Set-Theoretic Topology. – François G. Dorais♦ Aug 8 '12 at 16:53