Let $X$ be a compact Riemann surface and let $D$ be a divisor on $X$. With $1_D$ one generally indicates a particular meromorphic section of the invertible sheaf $\mathcal O_X(D)$ such that $$\operatorname{div }(1_D)=D$$
In other words $1_D$ is a collection of compatible meromorphic functions $\{f_\alpha\}$ on an open cover $\{V_\alpha\}$ of $X$ such that $$\operatorname{ord}_x (f_\alpha)=\operatorname{ord}_x(D)$$
I believe that this "collection" of meromorphic functions exists but how to construct the apparently distinguished one $1_D$? To be more precise: what is exactly $1_D$?
Example: On the stacks project I've found this definition for the algebraic case and when $D$ is effective, but honestly it is not clear to me.
One way of seeing this is "by direct definition:"
Define ${\cal L}={\cal O}_X(D)$ as the sub-sheaf of the constant sheaf ${\cal K}_X$ of meromorphic functions such that $$ {\cal L} (V) = \{ g \in {\cal K}_X \,| \ \text{div}|_V g + D|_V \ge 0 \}.$$
So a meromorphic section is any element of ${\cal K}_X$. We define $$\text{div}_{\cal L}\ g = D + \text{div}\ g,$$ where the div without the subscript is the usual div.
In particular, $\text{div}_{\cal L} (1) = D$.
Note - buyer beware - or feature - $$ \text{div}_{\cal L}\ f g = \text{div}\ f + \text{div}_{\cal L}\ g.$$
Alternatively - and equivalently - using compatible functions as you write above:
Assume that $D|_{V_\alpha} = \text{div }|_{V_\alpha} f_a$. Define ${\cal L}={\cal O}_X(D)$ by $$ {\cal L}|_{V_\alpha} = {1 \over {f_\alpha} } {\cal O}_X|_{V_\alpha} \subset {\cal K} .$$ This is well defined, because the $f_\alpha$ are unique up to multiplication by elements in ${\cal O}_X(V_\alpha)^*.$
If $s\in {\cal L}\otimes {\cal K} (X)$ is a global section (and so a meromorphic section of $\cal L$), where we write $ s|_{V_\alpha} = g_\alpha /f_\alpha$ with $g_\alpha \in {\cal K}$ locally, we define $ \text{div}_{\cal L} s$ by gluing (not adding!) the local definitions $$\text{div}_{\cal L}|_{V_\alpha}(s|_{V_\alpha}) = \text {div}|_{V_\alpha} g_\alpha.$$
In particular if $1 \in {\cal L}\otimes {\cal K} (X) $ is the section defined by the local equations $$1|_{V_\alpha} = { f_\alpha \over f_\alpha },$$ then again, $\text{div}_{\cal L} (1) = D$.
In the second approach, the expression $\{g_\alpha\}$ of a meromorphic section $s$ of $\cal L$ depends on the choice of $f_\alpha$ - but not its divisor $\text{div}_{\cal L} (s)$.
Explicit example, in the algebraic geometry setting: take $X= \mathbb P^1 $, with $U = \text { spec}\ k[t]$, and $ V = \text {spec}\ k[1/t]$. Take $D = n\cdot \infty$, with $f_U = 1$, and $f_V = 1/t^n$.
In the first approach, there's not much to say...
In the second approach, $$ {\cal L}(U) = k[t]\text {, and}\ {\cal L}(V)= 1/(1/t^n)k[1/t] = t^n k[1/t].$$
$$1_U = {1\over 1} \text{, and, }1 = t^n \cdot (1/t)^n,$$ so $\text {div}_{\cal L}|_U 1 = 0$, and $\text {div}_{\cal L}|_V 1 = \text {div}|_V (1/t)^n = n\cdot \infty$.
Edit/Addition: (To define $1_D$ explicitly, and to clear up some sloppiness above, taking especially into account the comment of Georges below.) The sheaf ${\cal L}$ is a subsheaf of the constant sheaf ${\cal K}_X$, so write $i\colon {\cal L} \to {\cal K}$ for the inclusion. In both approaches, I implicitly identified the domain with the codomain of the isomorphism $$ {\cal L}\otimes_{\cal O} {\cal K} \xrightarrow{i\otimes 1 }{\cal K}\otimes_{\cal O} {\cal K} \xrightarrow{\text{mult}}\cal K.$$ Under the identification, the well-defined, canonical meromorphic section $1_D$ of the original question is expressed locally (in the second approach) as $1/f_\alpha \otimes f_\alpha$ in the domain, and globally as $1$ in the codomain: $$ {1\over f_\alpha} \otimes f_\alpha \to {1\over f_\alpha} \otimes f_\alpha \to {1\over f_\alpha}\cdot f_\alpha = 1.$$