(Really a comment, but too long:)

If you strengthen your first question to demand that the subset actually be linearly ordered by $\prec$ (that is, strict chains) - which rules out, for example, Goldstern's construction - then the answer is different.

In the case $\kappa=\aleph_0$, we can divide the countable ordertypes into the *scattered* (does not embed $\eta$) and *unscattered* cases. Since unscattered linear orders all embed $\eta$, any countable linear order embeds into any countable unscattered order, so in any strict chain $C$, at most one unscattered order occurs, and if it occurs it occurs as the maximal element. So the question reduces to how long a strict chain of scattered orders can be.

Clearly there are strict chains of cardinality $\aleph_1$ (e.g., the countable ordinals). This answers the question fully, in case $2^{\aleph_0}=\aleph_1$, since there are exactly continuum many isomorphism types of countable linear orders. Now, I believe that this can be extended to the case where $CH$ fails - that is, no matter what the value of the continuum, strict chains of ordertypes under embeddability have cardinality $\le\aleph_1$ - as follows: there is a way of assigning a countable ordinal to a countable scattered linear order, called the *Hausdorff rank* of the linear order, and it can be shown (I believe) by induction on rank that any strict chain of (countable scattered) linear orders of bounded rank is countable. It then follows that the maximum cardinality of a strict chain of scattered countable linear orders (and hence countable linear orders) is $\aleph_1$. What this means, concretely, is that we can have universes of ZFC in which there are no strict chains of countable linear orders of length continuum.

Even better, this actually shows that for any strict chain $\mathcal{C}$, we have $$ \omega_1+2\not\prec \mathcal{C},$$ which is better than just a cardinality bound.

I have no idea what happens in the case $\kappa>\aleph_0$, if we look at strict chains.

CAVEAT: this is all just a sketch, and it may well be wrong; it's been a while since I've looked at this stuff.