The ordinals below $\omega^2$ are exactly those of the form $\omega\cdot n+k$ for natural numbers $n$ and $k$. Thus, these are the ordinals having two digits in base $\omega$, and counting to $\omega^2$ is much like counting to one hundred in this regard.
To order $\mathbb{N}$ in order type $\omega^2$, therefore, is essentially the same as to have a pairing function on the natural numbers, where $\langle n,k\rangle$ is the number that will be assigned to $\omega\cdot n+k$.
There are a huge variety of pairing functions, many of which are simpler than others by various measures.
My favorite is the Cantor pairing function $\langle n,k\rangle = \frac{(n+k)(n+k+1)}2 +k$, which has the advantage that it is a polynomial, easy to compute and invert, and has a simple definition.
Another simple pairing function is $\langle n,k\rangle = 2^n(2k+1)-1$, which has the advantage that it is easily seen to be bijective, and is also easy to compute, invert and define.
The pairing function that is implicit in your order, in contrast, is not so simple on the computational criteria, since it requires us to factor numbers into primes, which can be difficult.
Lastly, let me say that I don't find your question to be mathematically meaningful without a clearer criteria for simplicity or naturality. Shall we minimize the size of the Turing machine that computes the order? Shall we find a smallest defining arithmetic formula? Shall we minimize the logical complexity of the definition? Or what?
(This answer has been substantially edited since its original version.)
1.
The surreal number with sign-expansion $+-^{\omega}++-^{\omega}+++-^{\omega}++++-^{\omega}\cdots$ (indexed by $\omega^2$) can be rewritten in a tidy form, which shows that it does lie in the field generated by $\omega$.
As you read off that sign-expansion, the number follows the following pattern: $$1,1/2,1/4,1/8,\ldots$$
$$+-^{\omega}=\frac1\omega$$
$$\frac2\omega,\frac3\omega,\frac2\omega+\frac1{2\omega},\frac2\omega+\frac1{4\omega},\ldots$$
$$+-^{\omega}++-^{\omega}=\frac2\omega+\frac1{\omega^2}$$
$$+-^{\omega}++-^{\omega}+++-^{\omega}=\frac2\omega+\frac3{\omega^2}+\frac1{\omega^3}$$
So that we end up with $$\frac2\omega+\frac3{\omega^2}+\frac4{\omega^3}+\cdots=\frac{2\omega-1}{\left(\omega-1\right)^2}$$
Now, the fact that the finite sums and the subtler fact that the "infinite sum" works out in this way is not obvious.
2.
$\dfrac{2\omega-1}{\left(\omega-1\right)^2}$ is not actually a ratio of ordinals.
Treating the expression as a rational function of a real variable $\omega$, it diverges at $\omega=1$, but if it could be written as a rational function with positive coefficients (say by multiplying the numerator and denominator by a suitable polynomial), it could not diverge for any positive $\omega$. Incidentally, the inverse of this sort of argument is in Meissner's "Über positive Darstellungen von Polynomen" which contains the theorem that a polynomial which is positive for nonnegative real $x$ (such as $x^2-x+1$) can always be written as a rational function with positive coefficients (such as $(x^3+1)/(x+1)$).
3.
A sign expansion with the finite-orbit property need not be in the field generated by the ordinals.
$+^\omega-^{\omega^2}$ has the finite orbit property (indeed, the only nontrivial tails have the forms $+^\omega-^{\omega^2}$, $-^{\omega^2}$), but $+^\omega-^{\omega^2}=\sqrt{\omega}$ is not in the field generated by the ordinals (and hence certainly not a ratio of ordinals).
4.
A ratio of ordinals that are not both finite can still sometimes have the finite-orbit property. In particular, $\dfrac{1}{\omega}=+-^{\omega}$, and more interestingly, $\dfrac{1}{\omega+1}=\left(+-^{\omega}-+^{\omega}\right)^{\omega}$. The latter expansion has the finite orbit property, since the only forms of tails are $\left(+-^{\omega}-+^{\omega}\right)^{\omega}$, $-^{\omega}-+^{\omega}\left(+-^{\omega}-+^{\omega}\right)^{\omega}$, $-+^{\omega}\left(+-^{\omega}-+^{\omega}\right)^{\omega}$, and $+^{\omega}\left(+-^{\omega}-+^{\omega}\right)^{\omega}$.
$+-^{\omega}-=\dfrac{1}{\omega}-\dfrac{1}{\omega2}=\dfrac{1}{\omega2}$ Then for finite $n$, we have $+-^{\omega}-+^{n}=\dfrac{1}{\omega}-\dfrac{1}{\omega2}+\dfrac{1}{\omega4}+\cdots+\dfrac{1}{\omega2^{n-1}}=\dfrac{1}{\omega}-\dfrac{1}{\omega2^{n+1}}$. We end up with $+-^{\omega}-+^{\omega}=\dfrac{1}{\omega}-\dfrac{1}{\omega^2}$. Analogously, $\left(+-^{\omega}-+^{\omega}\right)^n=\displaystyle\sum_{k=1}^{2n}\dfrac{(-1)^{k-1}}{\omega^k}$. By the expansion of $\dfrac{1}{x+1}$ at infinity, we have $\dfrac{1}{\omega+1}=\left(+-^{\omega}-+^{\omega}\right)^\omega$.
5.
However, a ratio of ordinals need not have the finite-orbit property. In particular, $\dfrac{1}{\omega+2}=+-^{\omega}-+^{\omega}-^1+-^{\omega}+^3-+^\omega-^7+-^\omega+^{15}\cdots$. In a similar way to the analyses of $\dfrac{2\omega-1}{(\omega-1)^2}$ and $\dfrac{1}{\omega+1}$, this follows from the expansion at infinity: $\dfrac{1}{\omega+2}=\displaystyle\sum_{k=1}^{\infty}\dfrac{(-2)^{k-1}}{\omega^k}$.
Enough tools for justifying the above facts about sign expansions above are covered in VIII.2 and VIII.3 from Aaron N. Siegel's book Combinatorial Game Theory, but you can very likely also extract what you need from Philip Ehrlich's "Conway names, the simplicity hierarchy and the surreal number tree".
Best Answer
Alright, I'll put my comment as an answer and hopefully get this off the no-upvoted-answers queue. :)
Here's another nice-but-surprising way to get $\omega^\omega$: Let $\|n\|$ denote the smallest number of 1's needed to write n using any combination of addition and multiplication, e.g., $\|7\|=6$ as shortest way for 7 is $7=(1+1+1)(1+1)+1$. (This is known as the "integer complexity" of n; it's sequence A005245.)
Now, for any n, we have the lower bound $\|n\|\ge 3log_3 n$. So subtract this off and consider $\delta(n):=\|n\|-3log_3 n$. Then the set of all values of $\delta$ is a well-ordered subset of $\mathbb{R}$, with order type $\omega^\omega$.
For a proof, I refer you to my preprint: http://arxiv.org/abs/1310.2894