This is not an answer to all your questions, but anyway.
$\zeta_K(s)=\sum N(\mathfrak{a})^{-s}=\prod_\mathfrak{p} \frac{1}{1-N(\mathfrak{p})^{-s}}$. In the product we have $\frac{1}{1-p^{-s}}$ for every prime dividing $D$, $\frac{1}{(1-p^{-s})^2}$ for every prime that splits in $K$, i.e. such that $D$ is a square mod $p$, and $\frac{1}{1-p^{-2s}}= \frac{1}{(1-p^{-s})(1+p^{-s})}$ for every prime which doesn't split.
For every prime, the factor is thus $\frac{1}{1-p^{-s}}\frac{1}{1-(\frac{D}{p})p^{-s}}$. The product over all primes is thus $$\zeta(s)\,\prod_p \frac{1}{1-(\frac{D}{p})p^{-s}}$$ and your equation becomes
$$L(s,\chi)=\prod_p \frac{1}{1-(\frac{D}{p})p^{-s}}.$$
The function $\chi$ in $L(s,\chi)=\sum_n \chi(n) n^{-s}$ is a (quadratic) Dirichlet character modulo $D$, i.e. it is a group morphism $(\mathbb{Z}/D\mathbb{Z})^\times\to\{+1,-1\}$, which is then extended to a function $\mathbb{Z}\to \{0,+1,-1\}$ by $\chi(n)=0$ if $(n,D)\neq1$. Since $\chi$ is multiplicative, we have
$$L(s,\chi)=\prod_p \frac{1}{1-\chi(p)p^{-s}} .$$
We therefore indeed have $$\chi(p)=\left(\frac{D}{p}\right).$$
This implies that $(\frac{D}{p})$ depends on $p$ only modulo $D$ - something not evident at all from its definition, but an easy consequence of quadratic reciprocity. If in particular $D=(-1)^{(q-1)/2} q$ (where $q$ is a prime) then there is only one quadratic Dirichlet character, namely $\chi(n)=(\frac{n}{q})$. We therefore have
$$\left(\frac{p}{q}\right)=\left(\frac{(-1)^{(q-1)/2}\, q}{p}\right)$$
i.e. quadratic reciprocity.
Perhaps you'd be interested in a little more generality.
If $G$ is any finite group (say of order $n$), there are notions of representations of $G$. These are just homomorphisms $\rho:G\to\text{GL}(V)$ where $V$ is a f.d. $\mathbb{C}$-space. Such representations are called irreducible if the only non-trivial $\rho$-invariant (i.e. invariant under all the linear transformations in $\rho(G)\subseteq\text{GL}(V)$) subspaces of $V$ are the trivial ones.
Given any representation $\rho$ of $G$, one can associate to it a character $\chi_\rho:G\to \mathbb{C}$ defined by $\chi_\rho(g)=\text{tr}(\rho(g))$. The set $\text{irr}(G)$ of irreducible characters is merely the set of characters coming from an irreducible representation of $G$.
Now, while it may be non-obvious, the set $\text{irr}(G)$ is finite. In fact, a simplistic bound is that $\#\text{irr}(G)\leqslant n$, and in fact, $\#\text{irr}(G)$ is the number of conjugacy classes of $G$. This comes from the non-obvious fact that if $Z(\mathbb{C}[G])$ denotes the set of class functions on $G$ (i.e. functions $G\to\mathbb{C}$ which are constant on conjugacy classes) then $\text{irr}(G)$ forms a basis for $Z(\mathbb{C}[G])$. In fact, not only do they form a basis, they form an orthonormal basis. Of course, the obvious question, is with respect to what inner product? While it may seem opaque at first, if you haven't seen such matters before, the correct inner product on $Z(\mathbb{C}[G])$ is the weighted convolution:
$$\langle f_1,f_2\rangle=\frac{1}{|G|}\sum_{g\in G}f_1(g)\overline{f_2(g)}$$
Now, there is serious verbiage needed to justify why, in fact, the irreducible characters of $G$ form an orthornormal set with respect to this inner product. But, once this verbiage has been doled out, you get (by mere definition) the following equality
$$\frac{1}{|G|}\sum_{g\in G}\chi_i(g)\overline{\chi_j(g)}=\langle \chi_i,\chi_j\rangle=\delta_{i,j}$$
if $\text{irr}(G)=\{\chi_1,\ldots,\chi_m\}$. This is the so-called first orthogonality relation for irreducible characters.
The first in first orthgonality relation surely hints that we're not done--and we're not. There is a second orthogonality relation:
$$\frac{1}{|G|}\sum_{\chi\in\text{irr}(G)}\chi(g)\overline{\chi(h)}=\#(\mathbf{C}_G(g))c(g,h)$$
where $\mathbf{C}_G(g)$ denotes the centralizer of $g$ in $G$, and $c(g,h)$ is $1$ if $g$ is conjugate to $h$, and zero otherwise.
Now, at this point, you may be really confused as to what this has to do with your question. The answer is somewhat simple, but perhaps non-obvious. Note if $G$ is any group, and $\chi$ is a homomorphism $G\to\mathbb{C}^\times$, then in fact $\chi$ is a representation of $G$--since $\text{GL}_1(\mathbb{C})=\mathbb{C}^\times$. Moreover, since $\mathbb{C}$ has no non-trivial subspaces, such homomorphisms are trivially irreducible! Lastly noting that the trace of something in the range of $G\to\text{GL}_1(\mathbb{C})$ is nothing but the value that $\chi$ takes at the point of $g$, we can see that, in fact, $\text{Hom}(G,\mathbb{C}^\times)\subseteq\text{irr}(G)$. Now, I told you that $\#\text{irr}(G)$ is the number of conjugacy classes of $G$, and so when $G$ is abelian, $\#\text{irr}(G)=n$. But, if $G$ is abelian, say, you know that $\text{Hom}(G,\mathbb{C}^\times)\cong G$ (why?), and thus we can piece everything together to see that $\text{irr}(G)=\text{Hom}(G,\mathbb{C}^\times)$.
Thus, for an abelian group $G$, the orthogonality relations read as follows:
$$\frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\psi(g)}=\delta_{\chi,\psi}\qquad \chi,\psi\in\text{Hom}(G,\mathbb{C}^\times)\qquad\mathbf{(1)}$$
and
$$\frac{1}{|G|}\chi(g)\overline{\chi(h)}=|G|\delta_{g,h}\qquad \chi\in\text{Hom}(G,\mathbb{C}^\times)\qquad\mathbf{(2)}$$
Your two equalities are then special cases of these two identities. Indeed, for your first equality, let $\psi$ be the trivial character, i.e. the trivial map $G\to\mathbb{C}^\times$ in $\mathbf{(1)}$. For your second equality, let $h=1\in G$ in $\mathbf{(2)}$.
You can find proofs for the orthogonality relations in any good book on representation theory.
The above may not be of help to you, being at a possible opaque level of generality, but it's nice to put the theory of Dirichlet characters in the more general context of representation theory of finite groups. This elucidates "why" these identities occur, opposed to cute tricks like the proofs you gave above (which are totally fine, just more opaque). I hope this entices you to read more into the beautiful theory of representation theory. For a simple point of view, taking the attitude similar to what I talked about above, I would highly recommend Steinberg's book on the subject.
Best Answer
This is probably not a complete answer to your question but here it is anyways.
Usually, one calls a Dirichlet character $\chi$ quadratic, if it has order 2 in the dual of $(\mathbb{Z}/m\mathbb{Z})^\ast$, i.e. if $\chi^2 = 1$ and $\chi$ is not principal. Notice that every qudratic character takes only the values $\pm 1$ and $-1$ at least once. Equivalently, a quadratic characater is a real and non-principal character.
Next, call an integer $D$ a quadratic discriminant, if it arises as the discriminant of a quadratic number field. Equivalently, $D$ is a quadratic discrimant, if $D$ is $\neq 1$, square-free and $D \equiv 1 \mod{4}$ or if $D = 4d$ for some square-free integer $d \equiv 2$ or $3 \mod 4$.
Attached to every quadratic discriminant $D$ is a special quadratic character, called the Kronecker symbol, sometimes denoted by $\chi_D(n)$ or $\left( \frac{D}{n} \right)$ where $n$ is the argument. One can define it by requiring:
One has then the following Theorem, whose proof uses a lot of the theory of Dirichlet characters and Gauss' quadratic reciprocity law.
Theorem. For every quadratic discrimant $D$, the Kronecker symbol $\chi_D$ defines a primitive quadratic character mod $|D|$. Conversely, every primitve quadratic character is given by a Kronecker symbol and hence every quadratic character is induced by some Kronecker symbol.
In my opinion, Montgomery's and Vaughan's ''Multiplicative number theory'' or Davenport's classical ''Multiplicative number theory'' is a good reference for this.
Furthermore, the following formula holds for quadratic fields $K$:
$$\zeta_K(s) = \zeta(s)L(s, \chi_D).$$
In your notation $\chi_D = \chi_K$.
The answer to your first question is no with both your and 'my' defintion of quadratic.