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As you note, if we choose two different embeddings $k^s \to k_v^s$, say $\imath_1$ and $\imath_2$, then we get two different $G_{k_v}$-module structures on $M$, call them $M_1$ and $M_2$, and two different restriction maps $r_1:H^n(k,M) \to H^n(k_v,M_1)$ and $r_2:H^n(k,M) \to H^n(k_v,M_2)$.
The point is that there will also be a canonical isomorphism $i:H^n(k_v,M_1) \cong H^n(k_v,M_2)$ such that $i\circ r_1 = r_2,$ given by a formula analogous to the one you gave in the abelian variety context.

Namely, if $\imath_2 = \imath_1\circ g,$ then the isomorphism $i$ will be induced by $m \mapsto g\cdot m$ (if I have things straight; you can easily check if this is correct, of if I should have $g^{-1}$ instead). The fact that $i\circ r_1 = r_2$ will then depend on the fact that the automorphism of $H^n(k,M)$ induced by conjugation by $g^{-1}$ and the map $m\mapsto g\cdot m$ is the identity. So one does not have a literal independence of the embedding, but rather, the restriction is defined up to a canonical isomorphism independent of the embedding, and this latter fact does depend upon conjugation inducing the identity on cohomology (which is why people often summarize it in that way).

Note also that your abelian variety example is actually a special case of this, because the $M$ is $A(k)$, A(k^s)$, and the$G_{k_v}$-action on$M$does depend on the embedding of$k$k^s$ in $k_v$. k_v^s$. But the natural map$A(k) A(k^s) \to A(k_v)$A(k_v^s)$ also depends on this embedding, in such a way that, when you compose the restriction from $G_k$ to $G_{k_v}$ (with coefficients in $A(k)$) A(k^s)$) with the map on$G_{k_v}$-cohomology induced by the embedding$A(k) A(k^s) \hookrightarrow A(k_v)$A(k_v^s)$, you do obtain a map on cohomology that is independent of the embedding.

But it is not that in this case $M$ has a well-defined action of $G_{k_v}$ independent of the choice of embedding $k k^s \hookrightarrow k_v$k_v^s$. It is rather that$M$embeds into a bigger module$M_v$(in a way that also depends on the embedding) so that the composite$H^n(k,M)\to H^n(k_v,M) \to H^n(k_v,M_v)$is independent of the embedding. It is the embeddings$M \hookrightarrow M_v$that are missing in the more general context (i.e. when$M$is not of the form$A(k)$).A(k^s)$).

1

As you note, if we choose two different embeddings $k^s \to k_v^s$, say $\imath_1$ and $\imath_2$, then we get two different $G_{k_v}$-module structures on $M$, call them $M_1$ and $M_2$, and two different restriction maps $r_1:H^n(k,M) \to H^n(k_v,M_1)$ and $r_2:H^n(k,M) \to H^n(k_v,M_2)$.
The point is that there will also be a canonical isomorphism $i:H^n(k_v,M_1) \cong H^n(k_v,M_2)$ such that $i\circ r_1 = r_2,$ given by a formula analogous to the one you gave in the abelian variety context.

Namely, if $\imath_2 = \imath_1\circ g,$ then the isomorphism $i$ will be induced by $m \mapsto g\cdot m$ (if I have things straight; you can easily check if this is correct, of if I should have $g^{-1}$ instead). The fact that $i\circ r_1 = r_2$ will then depend on the fact that the automorphism of $H^n(k,M)$ induced by conjugation by $g^{-1}$ and the map $m\mapsto g\cdot m$ is the identity. So one does not have a literal independence of the embedding, but rather, the restriction is defined up to a canonical isomorphism independent of the embedding, and this latter fact does depend upon conjugation inducing the identity on cohomology (which is why people often summarize it in that way).

Note also that your abelian variety example is actually a special case of this, because the $M$ is $A(k)$, and the $G_{k_v}$-action on $M$ does depend on the embedding of $k$ in $k_v$. But the natural map $A(k) \to A(k_v)$ also depends on this embedding, in such a way that, when you compose the restriction from $G_k$ to $G_{k_v}$ (with coefficients in $A(k)$) with the map on $G_{k_v}$-cohomology induced by the embedding $A(k) \hookrightarrow A(k_v)$, you do obtain a map on cohomology that is independent of the embedding.

But it is not that in this case $M$ has a well-defined action of $G_{k_v}$ independent of the choice of embedding $k \hookrightarrow k_v$. It is rather that $M$ embeds into a bigger module $M_v$ (in a way that also depends on the embedding) so that the composite $H^n(k,M)\to H^n(k_v,M) \to H^n(k_v,M_v)$ is independent of the embedding. It is the embeddings $M \hookrightarrow M_v$ that are missing in the more general context (i.e. when $M$ is not of the form $A(k)$).
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