I think the confusion here mainly comes from the following misconception: For a number field $K$, infinite primes are not the same as embeddings $K\hookrightarrow\mathbb{C}$. They are rather equivalence classes of archimedean valuations on $K$. As Janusz proves, any such valuation is indeed induced by an embedding of $K$. However, if the embedding $\sigma$ is complex, then the conjugate embedding $\overline{\sigma}$ induces an equivalent valuation, i.e. the embeddings $\sigma$ and $\overline{\sigma}$ correspond to the same infinite prime! Thus, you have one infinite prime for each real embedding and one infinite prime for each conjugate pair of complex embeddings (note that indeed Janusz writes $\frak{P}_\text{$1$}, \dots, \frak{P}_\text{$r$}$ and $\frak{P}_\text{$r+1$}, \dots, \frak{P}_\text{$r+s$}$ in Prop. 2.3 for the infinite primes indicating that, indeed, you do not have $r+2s=n$ infinite primes, but only $r+s$).
Now for a relative extension $L/K$, let $\sigma$ be a real embedding of $K$ and let $\tau$ be a complex embedding of $L$ extending $\sigma$. Then the conjugate embedding $\overline{\tau}$ also extends $\sigma$! However, as we just saw, both $\tau$ and $\overline{\tau}$ correspond to the same infinite prime $\frak{P}$, which is why one can say that $\frak{P}$ has ramification index $2$ over the infinite prime $\frak{p}$ of $K$ corresponding to $\sigma$, i.e. $e_{\frak{P}|\frak{p}}=2$. If $\sigma$ is complex, this cannot occur. Here is another way to think about it: As $\tau=\mathrm{id}\circ\tau$ and $\overline{\tau}=(\overline{\tau}\circ\tau^{-1})\circ\tau$ are complex conjugate embeddings, they correspond to the same infinite prime $\frak{P}$ of $L$ and hence both $\mathrm{id}$ and $\overline{\tau}\circ\tau^{-1}\neq\mathrm{id}$ are in $D_\frak{P}$, i.e. $|D_\frak{P}\text{$|=2$}$.
Following this train of thought, one also gets a fundamental equality for infinite primes since any embedding of $K$ can be extended to an embedding of $L$ in $[L:K]$ ways. Identifying complex conjugate embeddings with the same infinite prime as explained above and setting all inertia degrees to one, one indeed obtains
$$
[L:K]=\sum_{\frak{P}|\frak{p}} e_{\frak{P}|\frak{p}}f_{\frak{P}|\frak{p}}.
$$
In particular, note that the number of infinite primes above the infinite primes $\frak{p}$ corresponding to the embedding $\sigma$ of $K$ is not, as you said, $n:=|\mathrm{Gal}(L/K)|$, but rather $r+s$, where $r$ is the number of real embeddings of $L$ extending $\sigma$ and $2s$ is the number of complex embeddings of $L$ extending $\sigma$.
In general, we can only have $e_{\frak{P}|\frak{p}}=1$ or $e_{\frak{P}|\frak{p}}=2$ for infinite primes $\frak{P}|\frak{p}$ and the ramification index can only be $2$ in the case outlined above, i.e. when $\frak{p}$ is real, but $\frak{P}$ is complex. The inertia degree $f_{\frak{P}|\frak{p}}$ is always $1$. (However, note that there are apparently authors who do this the other way around, i.e. they interpret the extension of a real infinite prime by a complex infinite prime as inertia instead of ramification and correspondingly, they always set $e_{\frak{P}|\frak{p}}=1$, while $f_{\frak{P}|\frak{p}}$ is either $1$ or $2$.)
As a concrete example, consider $K=\mathbb{Q}$ and $L=\mathbb{Q}(\sqrt{2})$. Then $L$ has two real infinite primes corresponding to $\sqrt{2}\mapsto\sqrt{2}$ and $\sqrt{2}\mapsto-\sqrt{2}$, which induce inequivalent archimedean valuations. Hence, they both have ramification index and inertia degree $1$ and you get the fundamental equality $1\cdot 1+1\cdot 1=2$.
If you consider instead $L=\mathbb{Q}(\sqrt{-2})$, then the (now complex!) embeddings $\sqrt{-2}\mapsto \sqrt{-2}$ and $\sqrt{-2}\mapsto -\sqrt{-2}$ are complex conjugates and hence induce the same valuation on $L$. Thus, $L$ has only one infinite prime, but with ramification index $2$ over the infinite prime of $\mathbb{Q}$ (this is precisely a case where a complex infinite prime lies over a real infinite prime as explained above) and the fundamental equality becomes $2\cdot 1=2$.
EDIT: Maybe one could go about Q2 in the following way. From the definition of ray class groups, we know that $a\equiv b\pmod{\frak{P}}$ for a real infinite prime $\frak{P}$ corresponding to the embedding $\sigma$ should mean $\sigma(a/b)>0$. If $\tau$ is another real embedding, then surely we would like to have the property that $a\equiv b\pmod{\frak{P}}$ implies $\tau(a)\equiv \tau(b)\pmod{\tau(\frak{P})}$ (similarly to the case of finite primes). If $\tau(\frak{P})$ corresponded to $\tau\circ\sigma$, then $(\tau\circ\sigma)(\tau(a)/\tau(b))=\tau\sigma\tau(a/b)$ and there is no reason why there should be any condition of the sign of this quantity. However, defining $\tau(\frak{P})$ to correspond to the embedding $\sigma\circ\tau^{-1}$, we find $(\sigma\circ\tau^{-1})(\tau(a)/\tau(b))=\sigma(a/b)>0$ so that, with this definition, the property $a\equiv b\pmod{\frak{P}}$ indeed implies $\tau(a)\equiv \tau(b)\pmod{\tau(\frak{P})}$.
Best Answer
Injectivity can be proven on "elements": (I'll be using notations as in Neukirch).
Assume that $\sigma \mathfrak{P}\cap L = \rho\mathfrak{P}\cap L = q$. Then, $\sigma \mathfrak{P}$ and $\rho \mathfrak{P}$ lie above $q$ in $N$. As $H$ acts transitively on the primes in $N$ that lie above $q$ (Proposition 9.1 in Neukrich), we get a $\tau \in H$: $\tau\sigma\mathfrak{P} = \rho \mathfrak{P}$.
Now, reformulating gives $\rho^{-1}\tau\sigma\mathfrak{P} = \mathfrak{P} \implies \rho^{-1}\tau\sigma \in G_{\mathfrak{P}}$. Thus, there exists $\varphi \in G_{\mathfrak{P}}$, such that $\rho = \tau\sigma\varphi^{-1}$. This proves that $\rho \in H\sigma G_{\mathfrak{P}}$ and thus $H\rho G_{\mathfrak{P}} = H\sigma G_{\mathfrak{P}}$.