Thank you Olivier for this great answer, I wish I could be one of your students ;-) !
Though I still not quite fully understand the adelic setting, here is some "classical" explanation I found, using the theory of ideals in orders of imaginary quadratic fields. I guess this is supposed to be a very well-known result but, since I haven't found any complete proof of it, I'll write down what I hope to be a detailed one in case some people are interested. My main reference is D.A. Cox 's "Primes of the form $x^2+ny^2$", chapters 7 and 8.
First, slightly modify the notations and set for any $f\in\mathbb{Z}$:
- $I_K(f)$: group of ideals of $\mathcal{O}_K$ which are prime to $f$ (i.e., whose norm is prime to $f$)
- $P_{K,\mathbb{Z}}(f)$: subgroup of $I_K(f)$ made of principal ideals of the form $\alpha\mathcal{O}_K$, with $\alpha$ congruent to some $a$ mod $f\mathcal{O}_K$, with $a\in\mathbb{Z}$ and $(a,f)=1$.
Here the ideals are meant to be integral ideals, and the group structure is obtained by identifying $\mathfrak{a}$ and $\mathfrak{b}$ iff there are some $\alpha$, $\beta\in \mathcal{O}_K$ such that $\alpha\mathfrak{a}=\beta\mathfrak{b}$.
The quotient $I_K(f)/P_{K,\mathbb{Z}}(f)$ is what we call a generalized ideal class group, isomorphic to $Pic(\mathcal{O}_f)\simeq \mathrm{Gal}(K_f/K)$ ($K_f/K $ is the ring class field of conductor $f$, the isomorphism being defined by sending any prime-to-$f$ prime ideal of $\mathcal{O}_K$ to its Frobenius element). Here we set $Pic(\mathcal{O}_f)$ to be the group formed by the quotient of proper ideals of $\mathcal{O}_f $ (i.e., ideals $\mathfrak{a}\subset \mathcal{O}_f$ such that $End(\mathfrak{a})=\mathcal{O}_f)$, by principal ideals. An important fact is that any (proper or not) $\mathcal{O}_f$-ideal is a free $\mathbb{Z}$-module of rank $2$.
The isomorphism between $I_K(f)/P_{K,\mathbb{Z}}(f)$ and $Pic(\mathcal{O}_f)$ is given by sending $$\mathcal{O}_K\supset\mathfrak{a} \mapsto \mathfrak{a}\cap\mathcal{O}_f$$
One checks that elements of $P_{K,\mathbb{Z}}(f)$, of the form $\alpha\mathcal{O}_K$ are sent to $\alpha\mathcal{O}_f $ (this follows from $(a,f)=1$, with $a$ "the" integer congruent to $\alpha$ mod $f\mathcal{O}_K$.) The inverse isomorphism is given by sending $$\mathcal{O}_f\supset\mathfrak{a}\mapsto \mathfrak{a}\mathcal{O}_K$$ (see Cox's 7.18, 7.20).
Though it's not strictly necessary, I mention that the norm is preserved by these maps. The case$ f=1$ gives us the Hilbert class field $K_1 $ of $K$, such that $I_K/P_K=Pic(\mathcal{O}_K)\simeq \mathrm{Gal}(K_1/K)$. With these identifications, one obtains that the group $\mathrm{Gal}(K_f/K_1)$ corresponds to $I_K(f)\cap P_K/P_{K,\mathbb{Z}}(f)$.
An important fact is that, if $\mathcal{O}_K^{\times}=\{\pm 1\}$, we have the following exact sequence: $$1\rightarrow (\mathbb{Z}/f\mathbb{Z})^{\times}\rightarrow (\mathcal{O}_K/f\mathcal{O}_K)^{\times}\rightarrow I_K(f)\cap P_K/P_{K,\mathbb{Z}}(f) \rightarrow 1$$ (see Cox, 7.27)
Applying this to $f=n$, $nl$, one gets that $$\mathrm{Gal}(K_{nl}/K_1)\simeq (\mathcal{O}_K/nl\mathcal{O}_K)^{\times}/(\mathbb{Z}/nl\mathbb{Z})^{\times}$$ $$\mathrm{Gal}(K_{n}/K_1)\simeq (\mathcal{O}_K/n\mathcal{O}_K)^{\times}/(\mathbb{Z}/n\mathbb{Z})^{\times}$$
Therefore $\mathrm{Gal}(K_{nl}/K_n)$ is isomorphic to $$\frac{(\mathcal{O}_K/nl\mathcal{O}_K)^{\times}/(\mathbb{Z}/nl\mathbb{Z})^{\times}}{(\mathcal{O}_K/n\mathcal{O}_K)^{\times}/(\mathbb{Z}/n\mathbb{Z})^{\times}} $$ $$\simeq (\mathcal{O}_K/l\mathcal{O}_K)^{\times}/(\mathbb{Z}/l\mathbb{Z})^{\times}$$ as $(n,l)=1$. This last group is isomorphic to $\mathbb{F}_{l^2}^{\times}/\mathbb{F}_{l}^{\times}$, for $l$ is inert in $K/\mathbb{Q}$, so has cardinal $l+1$.
Elements of the group $\mathrm{Gal}(K_{nl}/K_n)$ also corresponds to prime-to-$nl$ principal ideals of the form $\mathfrak{a}=\alpha\mathcal{O}_K$, with $\alpha$ congruent to $a\in\mathbb{Z}$ mod $n\mathcal{O}_K$, $(a,n)=1$. Notice that $\alpha\in\mathcal{O}_{nl}$ iff. it corresponds to the identity of $\mathrm{Gal}(K_{nl}/K_n)$. Denote by $\mathfrak{a}(\sigma)=\alpha(\sigma)\mathcal{O}_K$ such an ideal corresponding to $\sigma\in\mathrm{Gal}(K_{nl}/K_n)$, and set $\mathfrak{a}_{\sigma}:=\mathfrak{a}_{\sigma}\cap\mathcal{O}_{nl}$ the corresponding proper ideal of $\mathcal{O}_{nl}$.
As a consequence, if $\sigma\in\mathrm{Gal}(K_{nl}/K_n)$, then $\mathfrak{a}(\sigma)\cap\mathcal{O}_n=\alpha(\sigma)\mathcal{O}_n$ is principal, thus $\mathfrak{a}_{\sigma}\subset\mathcal{O}_{nl}$ satisfies $$\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma}\subset \mathcal{O}_n.$$ We have $$\frac{\mathcal{O}_n}{\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma}}\xrightarrow{\sim}\frac{\mathfrak{a}(\sigma)\cap\mathcal{O}_n}{\mathfrak{a}_{\sigma}}\hookrightarrow \frac{\mathcal{O}_n}{\mathcal{O}_{nl}},$$ the first map being multiplication by $\alpha(\sigma)$. As $\frac{\mathcal{O}_n}{\mathcal{O}_{nl}}\simeq \mathbb{Z}/l\mathbb{Z}$ and as $\frac{\mathfrak{a}(\sigma)\cap\mathcal{O}_n}{\mathfrak{a}_{\sigma}}$ is non-trivial (if $\sigma$ is not trivial then $\alpha(\sigma$) belongs to $\mathfrak{a}(\sigma)\cap\mathcal{O}_n$ but not to $\mathcal{O}_{nl})$; if $ \sigma$ is trivial then one directly gets $\frac{\mathfrak{a}(\sigma)\cap\mathcal{O}_n}{\mathfrak{a}_{\sigma}}=\frac{\mathcal{O}_{n}}{\mathcal{O}_{nl}})$, which finally implies that $$\frac{\mathcal{O}_n}{\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma}}\simeq \mathbb{Z}/l\mathbb{Z}$$
Combining this with the formula $$\mathrm{Tr}_{K_{nl}/K_n}(x_{nl})=\sum_{\sigma\in\mathrm{Gal}(K_{nl}/K_n)}[\mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\rightarrow \mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\mathcal{N}_{nl}^{-1}]$$ (see my first post), and the facts that 1- $End(\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma})=End(\mathfrak{a}_{\sigma})=\mathcal{O}_{nl}$, 2- the isomorphism class of an elliptic curve $\mathbb{C}/\Lambda$ is invariant under $\Lambda\mapsto \alpha\Lambda$ and 3- if $\sigma\neq \sigma' $ then $ \mathfrak{a}_{\sigma}$ and $\mathfrak{a}_{\sigma'}$ define different elements in $Pic(\mathcal{O}_{nl})$, one gets that $\mathrm{Tr}_{K_{nl}/K_n}(x_{nl})$ is a sum of $l+1$ distinct elements of the form $$[\mathbb{C}/\mathfrak{b}\rightarrow \mathbb{C}/\mathfrak{b}(\mathcal{N_n}\cap End(\mathfrak{b}))^{-1}]$$ with $\mathfrak{b}\subset \mathcal{O}_n$ a sub-lattice of index $l$. This is precisely the equality $$\mathrm{Tr}_{K_{nl}/K_n}(x_{nl})=T_l(x_n)$$ which I was looking for.