I am interested in how the nonstandard models "look like". I get the idea that the usual axioms describing successor and ordering don't preclude the existence of other chains other than the standard natural one, however only because I was told so — I fail to grasp exactly why this is the case.
For instance consider the following sentences,
(S1) $\forall x \exists y (Sx=y)$
(S2) $\forall x,y,z (Sx=y\to Sx=z \to y=z)$
(O1) $\forall x,y(Sx=y\to x<y)$
(O2) $\forall x,y,z(x<y\to y<z \to x<z)$
And most importantly,
(O3) $\forall x,y(x<y\vee y<x \vee x=y)$
I don't see how (O3) does not preclude the existence of points outside the main chain, since it asserts that all elements of the universe are pairwise comparable / the structure is connected — not? I know of the usual compactness argument applicable here though I would be interested in a more constructive way of introducing a nonstandard model for this set of sentences here so I could visualise it.
In this particular case, there is a very simple non-standard model. Let $X = \{0,1\} \times \mathbb N$ and let $<$ be the strict lexicographical order on $X$. Let $$ S \colon X \to X, (a,b) \mapsto (a,b+1). $$
It's easy to verify that $(X; S, <)$ is a model of (S1),(S2),(O1),(O2),(O3).
Intuitively $(X, <)$ is just $\mathbb N$ with an additional copy of $\mathbb N$ 'put on top'. So the 'non-standard' numbers in this case are of the form $(1,n)$ for $n \in \mathbb N$. And we have that every 'standard' numbers $(0,m)$ is below every non-standard number - a fact that holds true in general, but can be seen very explicitly in this model.