explicit formula for position of lowest bit of binary representation of number
As pointed out in the comments, $h(x)$ is the greatest $n$ such that $2^n$ divides $x$. This implies your observation that odd $x$ have $h(x)=0$ but also some semblance of periodicity in $h(x)$. Every second $h(x)$ is at least $1$, every fourth $h(x)$ is at least $2$ and so on. This is simply tally of how many twos divide $x$.
With this in mind, we can look for a function that varies periodically, according to the divisibility of its input. For example, we can choose the modulo operator, $\operatorname{mod}(x, y)$, which is the remainder of the Euclidean division of $x$ by $y$.
$\operatorname{mod}(x, y)$ rises from $0$ to a maximum of $(y-1)$. We can normalise it by $(y-1)$ and then take the floor of that, which gives us $1$ for every $y$'th number and $0$ otherwise. For example, $\left\lfloor\frac{\operatorname{mod}(x, 4)}{4-1}\right\rfloor$ is $1$ at every fourth integer and $0$ otherwise. The peak of the mod function is at $(y-1)$, so we then translate the previous function $1$ unit to the right so all $1$s are at multiples of $y$. Therefore $\left\lfloor\frac{\operatorname{mod}(x-1, y)}{y-1}\right\rfloor$ is the indicator of when $x$ is divisible by $y$. Summing these terms, when $y$ are different powers of two, gives a tally for the number of powers of two that divide $x$.
Hence, we have the explicit formula $$h(x)=\sum_{n=1}^\infty\left\lfloor\frac{\operatorname{mod}(x-1, 2^n)}{2^n-1}\right\rfloor$$
https://www.desmos.com/calculator/osiqmhxndi
You can also express the floor and modulo operators as series of sinusoids if you want a series composed solely of elementary functions.
Edit: Revisiting this, we can consider each $\left\lfloor\frac{\operatorname{mod}(x-1, y)}{y-1}\right\rfloor$ summand as a piecewise function $f_k(x)=\begin{cases}1,&0\leq x<1\\0,&1\leq x<2^k\end{cases}$, which is periodic on $\mathbb{R}$ with period $2^k$. The functions $f_k$ can be made into Fourier series by partitioning the relevant integrals over intervals according to the piecewise function and then solving them with elementary calculus. Lastly, we sum $f_k(x)$ over $k$, combine all $a_0$ terms of the Fourier series into the geometric series $\sum_{k=1}^\infty\frac{1}{2^k}=1$ and simplify with the angle sum formula for sine. This gives us the neat double series of elementary summands:
$$h(x)=1+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{\sin\left(cx\right)+\sin\left(c\left(1-x\right)\right)}{\pi n}$$
where $c=\frac{\pi n}{2^{k-1}}$. This numerically agrees with the previous formula. https://www.desmos.com/calculator/5hayvzzwil
Suggested by Gottfried.
In math, there is the $2$-adic valuation ... this $h$ is the reciprocal of it. $$ \nu_2(12) = \frac{1}{2} $$ And in general $\nu_2(2^k m) = 2^{-k}$ if $m$ is odd.