When do imaginary quadratic extensions have unique complex places? This question has been edited in light of some helpful comments from @AdamHughes below.
Let $F$
be a totally real number field,
i.e. $F=\mathbb{Q}(t)/m(t)$
where $m(t)\in\mathbb{Z}[t]$
and all roots of $m(t)$
lie in $\mathbb{R}$.
Let $K$
be a quadratic extension of $F$
so that $K=F(t)/(t^2+d)$
where $d\in F^+$.
Then $[K:F]=2$
and $[K:\mathbb{Q}]=2[F:\mathbb{Q}]$,
which also equals the number of embeddings of $K$
into $\mathbb{C}$.
The places of $K$
over the basefield $\mathbb{Q}$
are identified with these embeddings,
except that we count complex conjugates as the same place.
It seems to me that it is not possible for $K$
to embed into $\mathbb{R}$
because under any embedding, $d\in F$
is still required to have a negative square root.
This leads me to believe that all the places contributed by $[F:\mathbb{Q}]$
become non-real embeddings.
This is around where I feel like I am missing something,
so let me stop here.
The question is,
under what conditions
does $K$
have a unique complex place?
It seems to me that this would only be when $F=\mathbb{Q}$.
But that begs the question of why some of the books I'm reading talk about "an imaginary quadratic extension of a totally real number field, having a unique complex place," when they could just say "a quadratic field."
 A: This portion of the post is to address the revised question. For the original, scroll down.
The degree $[F(\sqrt{-d}):F]=2$ and since $F$ is totally real, $F\cap\Bbb Q(\sqrt{-d})=\Bbb Q$ shows that

$$\operatorname{Gal}(F(\sqrt{-d})/\Bbb Q)\cong\operatorname{Gal}(F/\Bbb Q)\times\operatorname{Gal}(\Bbb Q(\sqrt{-d})/\Bbb Q)$$

Now the field $\Bbb F(\sqrt{-d})$ has no real embeddings, since they symbol $\sqrt{-d}$ will always square to $-d$ no matter what, and so has $s=[F(\sqrt{-d}):\Bbb Q]=[F:\Bbb Q]$ complex embedding pairs, all of which are induced by the $s$ real-embeddings of $F$, in fact you can list out all the places of $F(\sqrt{-d})$ if you know that $\operatorname{Gal}(F/\Bbb Q)=\{\sigma_i\}_{i=1}^s$. They are
$$|\sigma_i(a)+\sigma_i(b)\sqrt{-d}|$$
where $|\cdot |$ is the absolute value given by any fixed choice of embedding of $F$ into $\Bbb C$. In particular if you look at the original answer, that makes sense since you chose the embedding of $F$ you only had the one, but when you let $F$ be more abstract, you get this nice listing instead. So the short answer is never unless $F=\Bbb Q$.

This is the original response
To the first question:  no:  the symbol $\sqrt{-d}\in F(\sqrt{-d})$ is defined by the property that $\sqrt{-d}^2=-d<0$ which cannot happen in a real field.
To the second, you want to note that since you have already fixed an embedded $F\subseteq \Bbb C$, you don't get any action there, your base field elements are fixed. So if say $F=\Bbb Q(\sqrt 2)$ where the symbol $\sqrt 2$is the unique positive real number squaring to $2$, then your embedding has $F$ fixed that way, the abstract field $\Bbb Q(\sqrt 2)\cong\Bbb Q[x]/(x^2-2)$ has multiple choices for the symbol, but you've said you've chosen the embedding already.
Now, what happens when I extend $F$? Well, $F(\sqrt{-d})$ has degree $1$or $2$ over $F$, and obviously we're assuming for simplicity that the degree is $2$, since $1$ is boring. Then what you're asking is really about $F[x]/(x^2+d)$, but the embeddings, and therefore the places are just $x\mapsto\pm\sqrt{-d}$ where here we overload the $\sqrt{-d}$ notation and mean it to be the unique upper-half plane element of $\Bbb C$ which squares to $-d$. So what you're asking about is $\operatorname{Gal}(F(\sqrt{-d})/F)$ which only has degree $2$.
So elements of this extension are of the form $a+b\sqrt{-d}$ where $a,b\in F$. Then the two possible absolute values are just

$$|a+b\sqrt{-d}|, |a-b\sqrt{-d}|$$

where $|\cdot |$ is the absolute value already present on $\Bbb C$ with $F$ as an embedded subfield. If $F\subseteq \Bbb R$ is real, then it's clear the two absolute values are complex conjugates of one another, hence induce equivalent places. In short:  they always induce a unique complex place.
