Linearly ordered sets "somewhat similar" to $\mathbb{Q}$ 
*

*$\pmb{\eta}$ - order type of $\mathbb{Q}$.

*$\pmb{\eta} + \pmb{1}$ - order type of $\mathbb{Q}\cap(0, 1]$.

*$\pmb{\omega_1}$ - order type of the first uncountable ordinal. 


Let's say that a linear order type is q-like if every proper initial segment of an instance of this type is order-isomorphic to $\pmb{\eta}$ or $\pmb{\eta}+\pmb{1}$. For example, $\pmb{\eta}$, $\pmb{\eta}+\pmb{1}$, $\pmb{\eta}\cdot\pmb{\omega_1}$ and $\pmb{\eta}+(\pmb{1}+\pmb{\eta})\cdot\pmb{\omega_1}$are q-like. 
Question: How many distinct q-like order types exist?
 A: This is a great question!
There are a large uncountable number of these orders.
Consider as a background order the long rational line, the order
$\mathcal{Q}=([0,1)\cap\mathbb{Q})\cdot\omega_1$. For each
$A\subset\omega_1$, let $\mathcal{Q}_A$ be the suborder obtained
by keeping the first point from the $\alpha^{\rm th}$ interval
whenever $\alpha\in A$, and omitting it if $\alpha\notin A$. That
is,
$$\mathcal{Q}_A=((0,1)\cap\mathbb{Q})\cdot\omega_1\ \cup\ \{(0,\alpha)\mid\alpha\in
A\}.$$
This corresponds to adding $\omega_1$ many copies of $\eta$ or
$1+\eta$, according to the pattern specified by $A$ as a subset of $\omega_1$.
I claim that the isomorphism types of these orders, except for the question of a least element, correspond precisely to
agreement-on-a-club for $A\subset\omega_1$.
Theorem. $\mathcal{Q}_A$ is isomorphic to $\mathcal{Q}_B$ 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.
Proof. If $A$ and $B$ agree on $0$ and agree on a club $C$, then we may build an
isomorphism between $\mathcal{Q}_A$ and $\mathcal{Q}_B$ by
transfinite induction. Namely, for each $\alpha\in C$, we will
ensure that $f$ restricted to the cut below $(0,\alpha)$ is an
isomorphism of the segment in $\mathcal{Q}_A$ to that in
$\mathcal{Q}_B$. The point is that $(0,\alpha)$ is actually a
point in $\mathcal{Q}_A$ if and only if it is point in
$\mathcal{Q}_B$, and so these points provide a common frame on
which to carry out the transfinite recursion. If we have an
isomorphism up to such a point, we can continue it to the next
point, since this is just adding a copy of $\eta$ or of $1+\eta$
on top of each (the same for each), and at limits we take the
union of what we have built so far, which still fulfills the
property because $C$ is closed. So $\mathcal{Q}_A\cong\mathcal{Q}_B$.
Conversely, if $f:\mathcal{Q}_A\cong\mathcal{Q}_B$, then $A$ and $B$ must agree on $0$. Let $C$ be
the set of closure ordinals of $f$, that is, the set of $\alpha$
such that $f$ respects the cut determined by the point
$(0,\alpha)$. This set $C$ is closed and unbounded in $\omega_1$.
Furthermore, it is now easy to see that $(0,\alpha)\in
\mathcal{Q}_A$ if and only if $(0,\alpha)\in \mathcal{Q}_B$ for
$\alpha\in C$, since this point is the supremum of that cut, and
it would have to be mapped to itself. Thus, $A\cap C=B\cap C$ and
so $A$ and $B$ agree modulo the club filter. QED
Corollary. There are $2^{\aleph_1}$ many distinct q-like linear orders up to isomorphism. 
Proof. The theorem shows that there are as many different q-like linear
orders as there are equivalence classes of subsets of $\omega_1$
modulo the non-stationary ideal. So the number of such orders is
$|P(\omega_1)/\text{NS}|$. This cardinality is $2^{\aleph_1}$
because we may split $\omega_1$ into $\omega_1$ many disjoint
stationary sets, by a theorem of Solovay and Ulam, and the union
of any two distinct subfamilies of these differ on a stationary
set and hence do not agree on a club.
So there are at least $2^{\aleph_1}$ many distinct q-like orders
up to isomorphism, and there cannot be more than this, since every
such order has cardinality at most $\omega_1$. QED
Finally, let me point out, as Joriki mentions in the comments, that every uncountable q-like linear order is isomorphic to $\mathcal{Q}_A$ for some $A$. If $L$ is any such order, then select an unbounded $\omega_1$ sequence in $L$, containing none of its limits, chop $L$ into the corresponding intervals these elements and define $A$ according to whether these corresponding intervals have a least element or not. Thus, we have a complete characterization of the q-like linear orders: the four countable orders, and then the orders $\mathcal{Q}_A$ with two $A$ from each class modulo the nonstationary ideal, one with $0$ and one without.
