Brauer group of a field and norms For $a,b \in K^\times$ the symbol $(a,b)$ denotes the element of the Brauer group of $K$ represented by the $2$-cocycle on the absolute Galois group $G_K$ of $K$ sending $(g_1,g_2)$ to 
$$
\sqrt{a}^{\varphi(g_1)+\varphi(g_2)-\varphi(g_1g_2)}
$$
where
$$
\varphi(g)= \begin{cases}
0 & \mathrm{if \ } g\sqrt{b}=\sqrt{b} \\
1 & \mathrm{if \ } g\sqrt{b}=-\sqrt{b}
\end{cases}
$$
By abuse of notation we let $(a,b)$ denote this $2$-cocycle.
Now $(a,b)=(b,a)$ and thus $(a,b)=1$ if and only if $a$ is a norm from $K(\sqrt{b})/K$ if and only if $b$ is a norm from $K(\sqrt{a})/K$.
Doing some calculations I saw that the possible values for $\varphi(g_1)+\varphi(g_2)-\varphi(g_1g_2)$ are either $0$ or $2$, so $(g_1,g_2)$ is sent either to $1$ or $a$ (it's sent to $a$ when $g_1 \sqrt{b}=g_2 \sqrt{b}=-\sqrt{b}$). 
Suppose now $a$ is a norm from $K(\sqrt{b})/K$ i.e. $a=x^2-by^2$ for some $x,y \in K$. Define now $\phi: G_K \to \overline{K}$ to be such that $\phi(g)=x+\sqrt{b}y$ if $g\sqrt{b}=-\sqrt{b}$ and $\phi(g)=1$ otherwise. Then we can see that 
$$
(a,b)(g_1,g_2)=\frac{g_1 \phi(g_2) \cdot \phi(g_1)}{\phi(g_1g_2)}
$$
i.e. $(a,b)$ is a $2$-coboundary.
Now suppose $\exists \ \phi : G_K \to \overline{K}$ such that 
$$
(a,b)(g_1,g_2)=\frac{g_1 \phi(g_2) \cdot \phi(g_1)}{\phi(g_1g_2)}
$$
Now $\mathrm{Gal}(K(\sqrt{b})/K)$ is generated by some $g \in G_K$ and we have 
$$
a=(a,b)(g,g)=\frac{g \phi(g) \cdot \phi(g)}{\phi(g^2)}=g \phi(g) \cdot \phi(g)
$$
and so $N_{K(\sqrt{b})/K)}(\phi(g))=a$
Is my reasoning correct?
 A: You seem to have modified your original question in which, if I remember well, the commutativity of the symbol $<a, b>$ was absent. But your definitions - of the Brauer group, of the symbol $<a,b>$ - are not precise enough. And besides, it does not appear in your proof that the equivalences between the 4 properties are not of the same nature. Let me try to explain.
1) Symmetry of the normic property, namely a is a norm from $K(\sqrt b)$ iff b is a norm from $K(\sqrt a)$. This is an elementary algebraic result : it is obvious that the quadratic form $X^2-aY^2-bZ^2$ represents $0$ iff $b$ is a norm from $K(\sqrt a)$; then it suffices to permute the variables $Y, Z$ .
2) (Anti)-commutativity of the symbol. There are at least two equivalent definitions of the Brauer group, via cohomology or via central simple algebras. Here the occurrence of cocycles with values $0,1$ seems to indicate that you opt for a cohomological definition, with additive notation for the values. But it is not a good idea to concentrate on calculations of cocycles, because after all the cohomological machinery functions with cohomology groups and their functorial properties. So, in the cohomological setting, the absolute Brauer group of $K$ is defined as $Br(K)=H^2(G_K, {K_{sep}}^*)$, which is the inductive limit of the relative Brauer groups $Br(L/K)=H^2(Gal(L/K), L^*)$ when $L$ runs through the finite Galois extensions of $K$.  
For $a,b$ living in $K^*$ or in $K^*/{K^*}^2$ (check that it doesn't matter), your symbol $<a,b>$ for $a,b \in K^*$ is then a certain cohomology class in $Br(K)$ defined as the image of $(a,b)$ under the cup-product pairing $H^1(G_K, {\mu}_2) \times H^1(G_K, {\mu}_2) \to H^2(G_K, {\mu}_2)$, with  ${\mu}_2=(\pm 1)$ (multiplicative notation for the values). Without detailing the cup-product, let us make more precise the definition of the various terms. In the cohomology of the exact sequence $1\to {\mu}_2\to {K_{sep}}^*\to {K_{sep}}^*\to 1$ (taking squares in ${K_{sep}}^*$ is a surjective homomorphism), Hilbert's thm.90 says that $H^1(G_K, {K_{sep}}^*)=0$, hence  $H^1(G_K, {\mu}_2)\cong K^*/{K^*}^2 $, as well as $H^2(G_K, {\mu}_2)\cong Br(K)[2] $ (= the subgroup of elements killed by $2$). Moreover, because $G_K$ acts trivially on ${\mu}_2, H^1(G_K, {\mu}_2)= Hom(G_K, {\mu}_2)$ and the latter Hom group is $\cong K^*/{K^*}^2$ by Kummer theory. So finally, the cup-product induces a  pairing $(K^*/{K^*}^2)\times (K^*/{K^*}^2)\to Br(K)[2], (a,b)\to <a,b>$.
NB: The general cup-product with values in $(\mu_n$) is anti-commutative, i.e. $<a,b>={<b,a>}^{-1}$. It is commutative here only because the coefficients are in $\mu_2$. Anyway, the anti-commutativity also does the job in the general case.
3) $a$ is a norm from $K(\sqrt b)$ iff$<a,b>=1$. The functorial approach has also the advantage to explain why and how are introduced some notions/definitions which sometimes give the impression to fall from the sky. For more clarity, I tackle the general case when $K$ contains the group $\mu_n$ of $n$-th roots of unity, with $n$ not divisible by char($K$). Assume given a degree $n$ cyclic extension $L=K(\sqrt [n](a)/K$ with Galois group $G$, and the Kummer isomorphism $\phi: G \cong Hom (<a>$ mod ${K^*}^n, \mu_n)$ [$\phi$ is the same as in your first line when $n=2$]. Then the map $K^*\to  H^2(G, K^*)$ sending an element $b\in K^*$ to the class of the cyclic algebra $(\phi, b)$ induces an isomorphism $H^2(G, L^*)\cong K^*/N(L^*)$, where $N$ is the norm map . It follows immediately that $b$ is a norm from $K(\sqrt a)$ iff the cyclic algebra $(\phi, b)$ is split (i.e. is isomorphic to the matrix algebra $M_2(K)$). I don't recall the definition of $(\phi, b)$, but suffice it to say that for $n=2$ it coincides with the generalized quaternion algebra $(a,b)$ with basis {$1,i,j,ij$} defined by $i^2=a, j^2=b, ij=-ji$. Thus the functorial approach not only explains the intervention of $\phi$, but it also establishes a link between the two definitions of $Br(K)$  ./.
Ref. The best seems to be the book "Central simple algebras and Galois cohomology" by P. Gille & T. Szamuely, Cambridge Univ. Press, chap. 1 and 4.
