Let $X$ be the set of functions $f:\mathbb{R}\rightarrow\mathbb{R}$ infinitely differentiable at $0$ with $f(0)=0$ and $f(x)>0$ when $x\neq 0$. Now let us define a partial order $\preceq$ on $X$ by saying that $f\preceq g$ if $\lim_{x\rightarrow 0}\frac{f}{g}=0$. (That is, if $f$ grows infinitely slower than $g$.) My question is, what is the order type of a maximal total suborder (AKA a maximal chain) of $X$ under $\preceq$?
This answer shows that any countable subset of $X$ has a lower bound. And I assume it can be similarly shown that any countable subset has an upper bound. So what kind of ordered set would a maximal total suborder of $X$ be order-isomorphic to? Would it be order isomorphic to $\omega_1^{*}+\omega_1$, or to the Long Line, or what?
[I will write $\prec$ for your $\preceq$, since this relation is a strict order.]
Let $Y$ consist of all $f\in X$ such that $f^{(n)}(0)=0$ for all $n$. Then $Y$ has the following saturation property:
I do not have the fortitude to write out the proof; it is a messy but essentially straightforward diagonalization. We have countably many constraints in constructing $c$ (for each $a\in A$ and each $n$ we must eventually have $c>na$, for each $b\in B$ and each $n$ we must eventually have $c<b/n$, and for each $n$ and $m$ we must eventually have $|c^{(n)}|<1/m$). We construct $c$ bit by bit as we approach $0$ from the right (or the left), and on each bit we add one new constraint that we are required to satisfy. Since there are countably many constraints, we can arrange that every constraint eventually gets added and so each constraint is satisfied in some neighborhood of $0$.
Now let us turn back to $X$. For $f\in X$, let $d(f)$ be the least $n$ such that $f^{(n)}(0)\neq 0$ (if no such $n$ exists, i.e. if $f\in Y$, we write $d(f)=\infty$). Since we require $f(x)>0$ for $x>0$, $d(f)$ must be even if it is finite. Moreover, observe that if $d(f)$ is finite, then for all $g$, $g\prec f$ iff $d(g)>d(f)$ (this is immediate from considering the degree $d(f)$ Taylor expansions with remainder of $f$ and $g$).
It follows that if $C\subset X$ is a maximal chain, then for each even positive integer $n$, $C$ contains exactly one element $f_n$ such that $d(f_n)=n$. So, $C$ has order-type $D+\omega^*$, where $D=C\cap Y$ is a maximal chain in $Y$.
By the saturation property of $Y$ above, $D$ must be countably saturated: for any countable $A,B\subseteq D$ with $a\prec b$ for all $a\in A$ and $b\in B$, there exists $c\in D$ such that $a\prec c\prec b$. So this says $D$ is fundamentally "uncountable" in a much stronger sense than any of the chains you mentioned: it is "uncountably dense". Not only is it a dense linear order, but every element must have uncountable cofinality from both sides. No element of $D$ can be approached above or below by a countable sequence, since you can always squeeze another element in between.
Assuming the continuum hypothesis, a back-and-forth argument shows that there is a unique countably saturated linear order of cardinality $\aleph_1=2^{\aleph_0}$ up to isomorphism. So in that case, $D$ is unique up to isomorphism and thus so is $C\cong D+\omega^*$.
In the absence of the continuum hypothesis, $D$ (and hence $C$) may not be unique up to isomorphism. In particular, it is consistent with ZFC for there to exist maximal chains $D,D'\subset Y$ which have different lower cofinalities (where by lower cofinality I mean the minimal cardinality of a subset that is unbounded below). This follows from a theorem of Hechler (described here); Hechler's theorem is about sequences of natural numbers but such a sequence $(a_n)$ can be turned into one of your smooth functions by taking $f(1/n)=1/a_n!$ for each $n$ and picking a nice smooth interpolation in between.