This sketch assumes $\mathtt{AC}$. I'll provide more details if necessary, but I think it's instructive to work through them.
One way to proceed is to consider three special cases, and then argue that in general you can reduce to some special case.
Case 1: The order leans to the right. That is, each point in $\Theta$ has only countably many points less than it. A straightforward transfinite induction allows you to build a copy of each countable ordinal $\alpha$ sitting inside $\Theta$ (at each stage, there are only countably many points to less than or equal to a point you've added, so you have plenty of ways to continue the construction).
Case 2: The order leans to the left. Predictably, this means each point in $\Theta$ sees countably many points greater than it. This is just like Case 1, except you build a reversed copy of $\alpha$.
EDIT: Asaf pointed out an oversight which has now been fixed (I hope!).
Case 3: The balanced case. For each $x \in \Theta$ the set of successors of $x$ does not fall into Case 1 and the set of predecessors of $x$ does not fall into Case 2. In this case you can by transfinite induction show directly that each $\alpha$ embeds into a balanced order. One approach here is to show that in each balanced order you can find an increasing sequence $(x_n)_{n \in \omega}$ such that for each $n$ the interval $(x_n, x_{n+1})$ is uncountable. If one of those intervals falls into Case 1 or 2 (which we've already handled), so we can assume they're all Case 3. This lets you squeeze ordinals you've already handled in sequence, letting you build bigger countable ordinals.
Intuitively, Case 1 should feel like you're working with an ordinal, Case 2 with a reverse ordinal, and Case 3 like the reals (but this is just a heuristic!).
So why are these cases sufficient?
Given an arbitrary $\Theta$, let $\Theta_l$ be the set of points with only countably many predecessors (think of this as the "left end" of $\Theta$), let $\Theta_r$ be those points with only countably many successors (the "right end"), and let $\Theta_m$ be the rest (the "middle"). If $\Theta_l$ is uncountable, you can apply Case 1 to it. Likewise with $\Theta_2$ and Case 2. So WLOG assume both are countable. Then, look for points $x \in \Theta_m$ such that the set of successors of $x$ does not fall into Case 1 and the set of predecessors of $x$ does not fall into Case 2. Show that if you can't find any, then $\Theta_m$ is balanced (be careful here!).