Let $G = \Gamma$, let $H$ be the decomposition group of $w|v$, so $H$ acts on $K_w$. Then
$$K \otimes_{k} k_v \simeq \mathrm{Ind}^{G}_{H} K_w,$$
and the same is true after passing to units. Hence, by Shapiro's Lemma,
$$H^2(G,(K \otimes_{k} k_v)^{\times}) \simeq H^2(H,K^{\times}_w)$$
But $H$ is identified with $\mathrm{Gal}(K_w/k_v)$, and the result you ask for is then a standard result in the computation of a Brauer group of a local field (Corollary 2, page 131, Serre's notes in Cassels-Frohlich). Note that Serre also shows that this group is generated by $u_{L/K} \in \mathrm{Br}(k_v)$ with invariant $1/d_v \in \mathbf{Q}/\mathbf{Z}$ ("canonically").
Actually, for a more direct reference, see Corollary 7.4(b) of Tate's notes in the next chapter of the same book on page 177.
As to the last question, the map is as you expect; viewing the first element as $1/d_v$ inside $\mathbf{Q}/\mathbf{Z}$ and the latter as $1/d$ inside $\mathbf{Q}/\mathbf{Z}$, it is induced by the identity map from $\mathbf{Q}/\mathbf{Z}$ to itself, so in your setting is the injective map sending $1/d_v$ to $(d/d_v) \cdot 1/d$. This is because $H^2(K/k,C_K)$ is more or less the $d$ torsion in $\mathbf{Q}/\mathbf{Z}$ which is the "final term" in the standard short exact sequence of class field theory:
$$0 \rightarrow \mathrm{Br}(k_v) \rightarrow \bigoplus_{v} \mathrm{Br}(k_v) \rightarrow \mathbf{Q}/\mathbf{Z} \rightarrow 0.$$
For a precise reference, combine the discussion in Consequence 9.6 of page 185 (of Tate's notes) with diagram (9) in section 11, bottom of page 195.
To make sure you look at the right edition of Cassels-Frohlich I am using the one with this page numbering:
https://www.math.arizona.edu/~cais/scans/Cassels-Frohlich-Algebraic_Number_Theory.pdf
Best Answer
In the Artin map $\newcommand{\Z}{\hat{\Bbb Z}}G_K\mapsto O_K^*\times\Z$ the induced map $G_K\to\Z$ corresponds to the action on the maximal unramified extension $H=K^{ur}$. In detail we have $$G_K\to\text{Gal}(K^{ur}/K)\cong\text{Gal}(k^{alg}/k)\cong\Z$$ where $k$ is the residue class field. By definition the inertia group consists of the elements of $G_K$ acting trivially on $K^{ur}$, so the image of the inertia group is contained in $O_K^*\times\{1\}$. Indeed it is equal to $O_K^*\times\{1\}$.