We know that $f(x) = \log(|\sin(x/2)|)$ is a $2\pi$ periodic function. Is there a well known function $g : \mathbb{R} \to \mathbb{R}$ (ideally decaying to zero at infinity) such that we can write $f$ as a periodisation of $g$, i.e. $$ f(x) = c + \sum_{n\in\mathbb{Z}}g(x+2\pi n)? $$ for some constant $c\in\mathbb{R}$. The closest I can seem to get is that $$ f(x) = -\log(2) + \frac{1}{2}\log(x^2) + \frac{1}{2}\sum_{n\in\mathbb{Z}\backslash\{0\}}\log\left(\frac{(x+2\pi n)^2}{(2\pi n)^2}\right). $$ which is not exactly the same. We also know that $$ f(x) = -\log(2) - \sum_{n=1}^{\infty}\frac{\cos(nx)}{n}, $$ but again, there is a similar problem.
Note: Following @ashepler's answer, I have added a preferred type for $g$, i.e. decaying to zero at infinity.
There are infinitely many choices for $g$. I do not know if they qualify as "well-known".
Strategy for generating $g$
We present one particular way to create such a function $g$. It is not the only way to do so (see below for the general characterization of such functions). Let $\varphi$ be a partition of the unity function, that is a function that verifies: $$\sum_{n\in\mathbb Z}\varphi(x+2\pi n) = 1\tag{1}$$ It's really easy to build such functions. You can use the indicator function of $(-\pi, \pi)$, a triangular function or even $\text{sinc}\left( \frac{x}{\pi}\right)$. You can also use any bump function $b(x)$ that doesn't vanish (e.g. a Gaussian) and build $$\varphi(x) = \frac{b(x)}{\sum_{n \in\mathbb Z}b(x+2\pi n)}$$
Then define $g$ of the following form: $$g(x)=\varphi(x)(f(x)-c)$$ Then it is easy to check that $$\sum_{n\in\mathbb Z}g(x+2\pi n)=\sum_{n\in\mathbb Z}\varphi(x+2\pi n)(f(x+2\pi n)-c)=f(x)\sum_{n\in\mathbb Z}\varphi(x+2\pi n)-c=f(x)-c$$ which gives you a solution. Note that I'm not sure if you really need any $c$, since you can always subtract it from $f$ and retain a periodic function (which is all you need).
Note also that this method allows you to build a solution $g$ that's arbitrarily smooth or with arbitrary decay at infinity.
Characterization of the general solutions
Let's try to identify the set of tempered distributions $g$ that satisfy $$f(x) = c + \sum_{n\in\mathbb{Z}}g(x+2\pi n)\tag{2}$$ Taking the Fourier transform: $$\begin{split} \hat f(\xi) &=c\delta(\xi) + \sum_{n\in\mathbb Z} \hat g(\xi)e^{4i\pi^2 \xi n}\\ &=c\delta(\xi) + \hat g(\xi)\sum_{n\in\mathbb Z} e^{4i\pi^2 \xi n}\\ &=c\delta(\xi) + \hat g(\xi)\sum_{n\in\mathbb Z} \delta\left(\xi-\frac n{2\pi}\right)\\ &=c\delta(\xi) + \sum_{n\in\mathbb Z} \hat g\left(\frac n{2\pi}\right)\delta\left(\xi-\frac n{2\pi}\right)\\ \end{split}$$ where we have used the Poisson summation formula to turn the sum of complex exponentials into a Dirac comb. We have also assumed that $\hat g$ is continuous at the multiples of $\frac 1 {2\pi}$, so that it is well-defined at those points. This is reasonable given that OP asked for $g$ to decay at infinity, so provided the decay is fast enough, continuity will be ensured.
On the other hand, as OP notes, $$f(x) = -\log(2) - \sum_{n=1}^{\infty}\frac{\cos(nx)}{n}\tag{3}$$ Taking, again, the Fourier transform: $$ \hat f(\xi) = -\log(2)\delta(\xi)-\sum_{n=1}^{+\infty}\frac 1 {2n}\left(\delta\left(\xi-\frac{n}{2\pi}\right)+\delta\left(\xi+\frac n {2\pi}\right)\right)\tag{4}$$ Identifying $(2)$ with $(4)$ yields the following charaterization of the set of solutions: $$\boxed{\left\{ \begin{split} c+\hat g(0)=-\log 2\\ \hat g(n)=-\frac 1 {2|n|} & \text{ if } n\neq 0 \end{split} \right.}\tag{5}$$
Conversely, if a tempered distribution $g$ verifies the equations above, then it can be periodized into $f$ according to $(2)$. This characterizes the solutions $g$ to the problem. As we said, it assumes that such $\hat g$ is a continuous function at the non-zero integers.
It is clear that there are an infinite number of solutions, since only $(5)$ only constrains the behavior on a discrete grid of points.
Final discussion
I initially considered the inverse Fourier transform of $\xi\mapsto -\frac 1 {2\pi}\frac 1 {2|\xi|}$ as a solution. Unfortunately, in that case $g(x)=\alpha\log|x|+\beta \delta(x)$, and doesn't decay at infinity.
Instead, observing that the constraints $(5)$ are only on the behavior of $\hat g$ at the multiples of $\frac 1 {2\pi}$, I looked for solutions $\hat g$ in the form of $$\hat g(\xi)=\sum_{n\in\mathbb Z}\hat g\left(\frac n{2\pi}\right)\hat\varphi\left(\xi-\frac n{2\pi}\right)=-(c+\log 2)\hat\varphi(\xi) -\sum_{n\neq 0}\frac 1 {2|n|}\hat\varphi\left(\xi-\frac n{2\pi}\right)$$ for a function $\hat \varphi$ that verifies: $$\hat\varphi\left(\frac n {2\pi}\right) =\delta_{n=0} \text{ Kronecker symbol}\tag{6}$$ Then I realized that this condition $(6)$ is equivalent to the partition of the unity $(1)$.
Finally, note that the approach proposed in this post doesn't assume anything on the form of $f$, only that it's periodic (you could use different Fourier coefficients in $(4)$ and obtain the corresponding characterization $(5)$.