Basic setup:
Let $F$ be a number field, $G = GL_{2,F}$ and $G' = SL_{2,F}$. I write $\mathbb{A}$ for the adeles over $F$ and $\mathbb{A}_f$ for the finite adeles. Similarly for ideles we write $\mathbb{I}$ and $\mathbb{I}_f$. For simplicity, let $A = \prod_{\mathfrak{p}} \mathcal{O}_{F_{\mathfrak{p}}}$, where $F_{\mathfrak{p}}$ is the $\mathfrak{p}$-adic completion of $F$ at $\mathfrak{p}$. Let $K_f = GL_2(A)$, and $K_{\infty}$ be the maximal compact connected subgroup mod center, i.e. $K_{\infty} = \prod_{\nu \mid \infty} K_{\nu}$, with $K_{\nu}$ being either $\mathbb{R}^* SO_2(\mathbb{R})$ or $\mathbb{C}^* U_2$. Similarly let $U_{\infty}$ be the groups without the centers, i.e. the product of $SO_2(\mathbb{R})$'s and $U_2$'s for real and complex places respectively, and $U_f = SL_2(A)$.
We have $$ X^1 = G'(F) \backslash G'(\mathbb{A}) / U_f U_{\infty} \hookrightarrow G(F) \backslash G(\mathbb{A}) / K_f K_{\infty} = X $$ induced by the inclusion of $G'$ to $G$.
Further we have the determinant map $$ X = G(F) \backslash G(\mathbb{A}) / K_f K_{\infty} \to \mathbb{I} / F^* A^* F_{\infty}^* = \mathbb{I}_f / F^* A^*. $$ This latter group is (isomorphic to) the ideal class group of $F$.
It is often mentioned that the connected components of $X$ are in one to one correspondence with the elements of $\mathbb{I} / F^* A^*$, i.e. with the class group. It is said that this follows from strong approximation for $SL_2$.
The question is now the following: How?
My idea is that $X^1$ is the fibre over $1 \in \mathbb{I}_f / F^* A^*$ and connected. As the determinant is surjective and the fibres are isomorphic, the connected components of $X$ and $\mathbb{I}_f/F^*A^*$ agree. The latter is a finite discrete group, so the statement.
I believe, I am using the strong approximation propertry of $G'$ when I say, that $X^1$ is connected, but I don't really see this.
Note: For me, strong approximation for an algebraic group $H$ over $F$ is that $H(F) \subseteq H(\mathbb{A})$ is dense.
Final remark: As it might be unclear, I want to stress that I would like to get an opinion on my idea of the proof and hints or proves of the missing details.
Edit: The question can be generalized for arbitrary compact open subgroups of $G(\mathbb{A})$ instead of $K_f$. I believe, the idea of the proof would be essentially the same. The class group then becomes a class group with respect to a conductor.
Most of this should be in https://www.jmilne.org/math/xnotes/svi.pdf, but let me at least answer one of your questions here:
First of all note that $G'(\mathbb{R})$ is connected, so that also $G'(\mathbb{R})/K_{\infty}$ is connected, so it suffices to determine the connected components of $G'(F)\backslash G'(\mathbb{A}_f)/K_f$. Strong approximation for $G'(F)$ tells us that $G'(F)$ is dense in $G(\mathbb{A}_f)$, which implies that $G'(\mathbb{Q}) K_f = G(\mathbb{A}_f)$. Indeed, $G(\mathbb{A}_f)/K_f$ has the discrete topology and the image of $G'(F)$ in $G(\mathbb{A}_f)/K_f$ is dense, so it must be everything.