How do I double the resolution of $2^{\nu_2(x)}$? 
How do I double the resolution of $2^{\nu_2(x)}$?

$f(x)=2^{\nu_2(x)}$ gives the highest power of $2$ that divides $x$, e.g. $2^{\nu_2(28)}=4$
And this can measure consistently over $\Bbb N, \Bbb Z, \Bbb Z[\frac12], \Bbb Z[\frac16], \Bbb Z_2, \Bbb Q_2$ etc.
As it stands this only measures integer powers of $2$, but I want to "double" its resolution so it measures down to $\sqrt2$ in a consistent way.  How do I do this, and what sets does the new function range over consistently?

My thinking is that the answer is to set:
$\displaystyle \large {g(x)=(\sqrt2)^{\nu_2(x^2)}}$
a) does that do it, and
b) what analogues of the above list of sets does this function range over consistently?
In respect of part b), I mean, for example: instead of $\Bbb Z[\frac12]$ it might measure over $\Bbb Z\left[\frac1{\sqrt2}\right]$, and is the completion of $\Bbb Z$ or $\Bbb Q$ with respect to this metric different to $\Bbb Z_2$ and $\Bbb Q_2$?
 A: To quote from an answer I gave to one of your questions six months ago (but with new emphasis added):

As long as a number $\alpha$ is algebraic over $\mathbb{Q}$, there is a natural extension of the $2$-adic absolute value both from $\mathbb{Q}$ to $\mathbb{Q}(\alpha)$ and from $\mathbb{Q}_2$ to $\mathbb{Q}_2(\alpha)$. (EDIT: ... and up to $[\Bbb Q(\alpha):\Bbb Q]$ reasonable extensions of $|\cdot|_2$ to $\Bbb Q(\alpha)$, namely, one for each non-archimedean place of $\Bbb Q(\alpha)$ lying above $2$ ($\Leftrightarrow$ one for each irreducible factor of the $\Bbb Q$-minimal polynomial of $\alpha$ when viewed over $\Bbb Q_2$ $\Leftrightarrow$ one for each distinct prime factor in the prime factorisation of $(2)$ in the ring of integers of $\Bbb Q(\alpha)$. In the example $\alpha=\sqrt{2}$, there is only one such extension of the value, as the prime 2 totally ramifies.) 
As in your example, of course the only reasonable extension of the absolute value to $\alpha = \sqrt c$ for $c \in \mathbb{Q}_2$ is $|\sqrt c| := \sqrt{|c|}$. Much more generally, an element $\alpha$ whose minimal polynomial over $\mathbb{Q}_2$ has degree $d$ is given the value $|N_{\mathbb{Q}_2(\alpha)|\mathbb{Q}_2}(\alpha)|^{1/d}$. This is treated extensively in any book on $p$-adics, local fields etc.

With basic knowledge of $p$-adic (multiplicative) absolute values $|\cdot|_p$ and (additive) valuations $v_p$, the last definition reads
$v_2(\alpha) := \frac{1}{d}v_2(N_{\mathbb{Q}_2(\alpha)|\mathbb{Q}_2}(\alpha))$
or in your extremely special case
$v_2(\sqrt{2}) = \frac{1}{2}$
which uniquely defines an extension of $v_2$ to $\Bbb Q(\sqrt2)$ (or rather to the much bigger $\Bbb Q_2(\sqrt2)$), namely, as Somos' answer also states, ,
$$\tilde{\nu}_2(a+b\sqrt{2}) = \min (\nu_2(a), \nu_2(b) + \frac{1}{2}) \qquad \text{for } \qquad a,b \in \Bbb Q \, \text{ (or } \Bbb Q_2 \text{)}.$$
What any good lecture notes, books etc. treat is in what precise sense this extension of the value/valuation is unique, how it leads to the theory of ramified and unramified extensions etc.
On a personal note, I doubt that you followed the lead in the last sentence back then, I doubt you follow it now, and I assume you will ask a similar question in six, twelve, eighteen months again.
A: I think what you want already partially exists. Quoting the Wikipedia article p-adic order:

In number theory, for a given prime number $p$, the $p$-adic order or
  $p$-adic valuation of a non-zero integer $n$ is the highest exponent $\nu$
  such that $p^\nu$ divides $n$.

The article goes on to extend the function

The $p$-adic order can be extended into the rational numbers. We can define
  $\nu_p: \mathbb{Q}\to \mathbb{Z}$
  $$ \nu_p(a/b) = \nu_p(a)-\nu_p(b).$$

All you need to do is extend one step further. There is a severe restriction though. You have to define the domain of the function. It should be a field and the extended valuation should have properties that extend those it already has.
Given a field, there are field extensions. For example, $\mathbb{Q}(\sqrt{2}).$ You can extend the valuation $\nu_2$ to the extended field. Thus
 $\, \nu_2(x\sqrt{2}) = \frac12 + \nu_2(x) \,$ if
 $\, x \in \mathbb{Q}^\times. \,$ Also, if
 $\, x, \, y \in \mathbb{Q}^\times, \, $ then
 $\, \nu_2(x + y\sqrt{2}) = \min(\nu_2(x),\nu_2(y\sqrt{2})), \,$ as usual.
