I don't know of a reference, but the duality in question can be proved by results from Brown's book on group cohomology. I'll show the case $s\ge 0$. First note that for each integer $j$ there is an isomorphism

$$\psi: \hat{H}^j(G,k) \xrightarrow{\sim} \text{Hom}_k(\hat{H}_j(G,k),k)$$

(Brown, VI.7.2) and for $s\ge 0$ there is an isomorphism $$\varphi: \hat{H}_{-s-1}(G,k) \xrightarrow{\sim} H^s(G,k)$$

(Brown, VI.4). Denote the $k$-dual of a vector space or of a homomorphism by $(-)^\ast$. Tate duality is then the composition
$$t = (\varphi^{-1})^\ast\circ \psi: \hat{H}^{-s-1}(G,k) \xrightarrow{\sim} \hat{H}_{-s-1}(G,k)^\ast \xrightarrow{\sim} H^s(G,k)^\ast.$$
Hence we have to show the commutativity of the diagramm

$$\begin{array}{ccccc}
\hat{H}^{-s-1}(G,k) & \xrightarrow{\psi} & \hat{H}_{-s-1}(G,k)^\ast & \xleftarrow{\varphi^\ast} & H^s(G,k)^\ast \newline
{\scriptstyle \widehat{res}} \downarrow & & \downarrow \scriptstyle res^\ast & & \downarrow \scriptstyle tr^\ast\newline
\hat{H}^{-s-1}(H,k) & \xrightarrow{\psi} & \hat{H}_t(H,k)^\ast & \xleftarrow{\varphi^\ast} & H^s(H,k)^\ast
\end{array}$$

($t$ stands for $-s-1$ which the editor doesn't accept!?) The commutativity of the left hand square follows right from the definition of the maps and the right hand square commutes if we can show the commutativity of the following square:
$$\begin{array}{ccc}
\hat{H}_{-s-1}(G,k) & \xrightarrow{\varphi} & H^s(G,k)\newline
{\scriptstyle \widehat{res}} \uparrow & & \uparrow \scriptstyle tr^G_H \newline
\hat{H}_{-s-1}(H,k) & \xrightarrow{\varphi} & H^s(H,k)
\end{array}\tag{1}$$
In order to describe $\varphi$ on chain level, let $P \to k$ be a projective resolution over $kG$ and let $F$ be a complete resolution such that $F_i=P_i$ and $F_{-i-1} = P_i^\ast$ for $i \ge 0$. Then $\varphi$ is induced by the composition
$$\varphi: F_{-i-1}\otimes_{kG}k=P_i^\ast \otimes_{kG}k \xrightarrow{\alpha\otimes id} \text{Hom}_{kG}(P_i,kG) \otimes_{kG} k\xrightarrow{\beta}\text{Hom}_{kG}(P_i,k)$$
where $\alpha(f)(x)=\sum_{g \in G}f(g^{-1}x)g$ (Brown, VI.3.4) and $\beta(f \otimes a)(x)=f(x)a$ (Brown, I.8.3). Hence
$$\varphi(f \otimes a)(x)=\sum_{g \in G}f(g^{-1}x)(ga)=\sum_{g \in G}f(g^{-1}x)a=tr^G_E(f)(x)a\tag{2}$$
where $f \in P_i^\ast, a \in k, x \in P_i$ and $E=\{1\}$.

On chain level $(1)$ is given by the diagramm
$$\begin{array}{ccc}
P_i^\ast \otimes_{kG} k & \xrightarrow{\varphi_G} & \text{Hom}_{kG}(P_i,k) \newline
{\scriptstyle \kappa} \uparrow & & \uparrow \scriptstyle tr^G_H \newline
P_i^\ast \otimes_{kH} k & \xrightarrow[\varphi_H]{} & \text{Hom}_{kH}(P_i,k) \newline
\end{array}\tag{3}$$

where $\kappa(f \otimes_H a)=f \otimes_G a$. With $f,a,x$ as above, we obtain
$$(tr^G_H \circ \varphi_H)(f \otimes_H a)(x)=\sum_{g \in G/H}\varphi_H(f\otimes_H a)(g^{-1}x)
\overset{(2)}{=}\sum_{g \in G/H}tr^H_E(f)(g^{-1}x)a$$
$$\qquad=tr^G_H(tr^H_E(f))(x)a=tr^G_E(f)(x)a$$
$$\qquad\qquad=\varphi_G(f \otimes_G a)(x)=(\varphi_G \circ res)(f \otimes_H a)(x)$$
Thus the commutativity of $(3)$ is shown. **QED**