I'm looking for an explanation or a reference on why rank $n$ vector bundles on a curve over a field can be described as $\mathrm{GL}_n(\mathcal O)\backslash \mathrm{GL}_n(\Bbb A)/\mathrm{GL}_n(K)$, where $K$ is the function field of the curve, $\Bbb A$ is the corresponding ring of adeles and $\mathcal O\subset \Bbb A$ is the subring of integral adeles.
I'm aware of this question and its answers, but it seems overkill for this special case.
Let $V$ be a rank $n$ vector bundle over $X$. Then $V$ is trivial on a dense open subset $U=X\backslash S$ of $X$. Let $f\colon V\to \mathcal O_X^n$ be a meromorphic isomorphism (i.e., an isomorphism on $U$). Now for each $x\in X$ the map $f$ gives rise to a $\mathcal O_x$-linear map $V_x\hookrightarrow K_x^n$ which realizes $V_x$ as a $\mathcal O_x$-lattice of rank $n$. This can be viewed as an element $[V_x]\in\mathrm{GL}_n(K_x)/\mathrm{GL}_n(\mathcal O_x)$. Moreover, for $x\in U$ this element is trivial. This exactly defines an element in $\prod_{x\in X}[V_x]\in\mathrm{GL}_n(\mathbb A)/\mathrm{GL}_n(\mathcal O)$. A different choice of the meromorphic isomorphism differs by an element $A\in\mathrm{GL}_n(K)$, so we actually obtain an element of $\mathrm{GL}_n(K)\backslash \mathrm{GL}_n(\mathbb A)/\mathrm{GL}_n(\mathcal O)$.
Conversely, given an element of $\mathrm{GL}_n(K)\backslash \mathrm{GL}_n(\mathbb A)/\mathrm{GL}_n(\mathcal O)$, by going back along the construction of the last paragraph, we obtain, for each $x\in X$, a $\mathcal O_x$-lattice $V_x\subset K_x^n$, such that for $x$ in a dense subset, the lattice $V_x$ is the standard one $\mathcal O_x^n$. Now, we can obtain a coherent vector bundle $V $ whose sections at $U\subset X$ is $$\{f\in K^n:\forall x\in U,f_x\in V_x\}.$$ One can readily check that this defines a locally free $\mathcal O_X$-module of rank $n$ on $X$.
I will elaborate on what I mean by a meromorphic isomorphism, since I was asked to in the comments.
Let $V$ be be a vector bundle trivial on a dense open subset $U$ of $X$. Then we have an isomorphism $f\colon V|_U\to \mathcal O_U^n$. In particular, this induces an isomorphism $V_\zeta\simeq K^n$, where $V_\zeta$ is the stalk of $V$ at the generic point $\zeta$ of $X$, which is a $K=k(X)$-vector space of rank $n$. Now, there is a map $V_x\hookrightarrow V_\zeta\otimes_{k(X)}K_x\simeq K_x^n$, which is what I use in the rest of the answer.