What does it mean for the surreal numbers/partizan games to be "universally embedding"? In "On numbers and games", Conway writes that the surreal Numbers form a universally embedding totally ordered Field. Later Jacob Lurie proved that (the equivalence classes of) the partizan games form a universally embedding partially ordered abelian group.
As far as I can tell from Lurie's paper, the latter means: 
Let $\mathbb{U}$ denote the class of equivalence classes of partizan games. Furthermore let $S \subseteq S'$ be partially ordered abelian groups. Suppose that $\phi: S \rightarrow \mathbb{U}$ is an order-preserving homomorphism. Then there exists an order-preserving homomorphism $\phi': S' \rightarrow \mathbb{U}$ such that $\phi' \mid S = \phi$.
Now the thing is: I had before gotten to the (informal) conception that "universally embedding" in this context means that every (set-sized) partially ordered abelian group is isomorphic to some subgroup of $\mathbb{U}$, and vice versa that every totally ordered field is isomorphic to some subfield of $\mathbf{No}$. My question now is: is that even true? And what does this have to do with the statement of Lurie's paper?
 A: Besides the two types of structures being considered, there are two types of universality here. They are indeed related, but the stronger one (a la Lurie and I believe also Conway) is in my opinion the "right" one.
The story I like to imagine is this. We have some putatively-"universal" object $\mathcal{U}$ that we understand well, and some mysterious object $\mathcal{A}$ of the same type. Certainly at a minimum we should be able to find an embedding $\varphi$ of $\mathcal{A}$ into $\mathcal{U}$. However, what if we subsequently find out that $\mathcal{A}$ itself was really just a tiny fragment of some larger structure $\mathcal{B}$? We'd not only like to be able to embed $\mathcal{B}$ into $\mathcal{U}$, we'd also like to not have to undo the work we've already done: we want to find an embedding $\psi:\mathcal{B}\rightarrow\mathcal{U}$ which extends the already-thought-up embedding $\varphi$. (One way this story can become literally true is if we're trying to show that the "$\forall\exists$-theory" of $\mathcal{U}$ is reasonably simple, which happens e.g. in computability theory for various partial orders.)
To see an example of how this can play out, consider $\mathbb{Z}$ and $\mathbb{Q}$ as linear orders. Obviously each is "weakly universal" for all finite linear orders: any finite linear order embeds into both $\mathbb{Z}$ and $\mathbb{Q}$. However, only $\mathbb{Q}$ is "strongly universal" for finite linear orders: letting $\mathbf k$ denote the usual linear order with underlying set $\{1,\dotsc,k\}$, given any embedding $\varphi:\mathbf 2\rightarrow\mathbb{Z}$ there is some finite linear order $\mathbf n\supseteq\mathbf 2$ such that no extension of $\varphi$ embeds $\mathbf n$ into $\mathbb{Z}$ (basically just take $n$ much greater than $\varphi(2)-\varphi(1)$). By contrast, the usual "back-and-forth" argument for $\mathbb{Q}$ shows that $\mathbb{Q}$ is not subject to this sort of silliness. Note that the stronger type of universality brings with it a sort of "homogeneity" — a relevant  notion here is Fraïssé limit.
