Well, although *there is a typo* (Wiles forgot to close his parenthesis, and wrote $H^2(G,\mu_{p^r}\to H^2(G,\mu_{p^s})$ in his proof), his claim is correct. Let, as *ibid.* $F$ be the finite extension of $\mathbb{Q}_p$ fixed by $G$, so that your arrow can be written $H^2(F,\mu_{p^r})\to H^2(F,\mu_{p^s})$. Consider the Kummer sequence (for any $n\geq 1$)
$$
1\longrightarrow \mu_{p^n}\longrightarrow \overline{F}^\times\overset{(\cdot)^{ p^n}}{\longrightarrow}\overline{F}^\times\longrightarrow 1.
$$
Taking Galois cohomology and using Hilbert '90, which tells you $H^1(F,\overline{F}^\times)=1$, you find
$$
1\longrightarrow H^2(F,\mu_{p^n})\longrightarrow H^2(F,\overline{F}^\times)\overset{\cdot p^n}{\longrightarrow }H^2(F,\overline{F}^\times).
$$
Now, local class field theory tells you that the Brauer group $H^2(F,\overline{F}^\times)$ is isomorphic to $\mathbb{Q}/\mathbb{Z}$, so the above sequence identifies $H^2(F,\mu_{p^n})$ with the group $\mathbb{Z}/p^n$, seen as the kernel of multiplication by $p^n$ on $\mathbb{Q}/\mathbb{Z}$. If now you apply this for $r\leq s$, you see that the arrow you were firstly interested in is the injection $\mathbb{Z}/p^r\hookrightarrow \mathbb{Z}/p^s$ or, if you prefer, the injection
$$
\Big(\ker(\cdot p^r)\hookrightarrow \ker(\cdot p^s)\Big)\subseteq \mathbb{Q}/\mathbb{Z}.
$$

**Added Jan 24th** Concerning the second exact sequence (that involving $H^1$), *there, there is a typo!* What Wiles wanted to write was that the natural map
$$\tag{1}
H^1(G,\mu_{p^n})\otimes M\longrightarrow H^1(G,M(1))
$$
is an isomorphism, but in his paper he made (again!) a mistake with parenthesis and *ibid* he replaces the first term by $H^1(G,\mu_{p^n}\otimes M)$. You can see (1) in the quoted paper by Diamond and, moreover, it is (1) which shows up later in Wiles' proof: indeed, he deduces from (1) something about another sequence where $H^1(G,\mu_{p^n})\otimes M$ is replaced, by Kummer theory, by $((F^\times/(F^\times)^{p^n})\otimes M$, which he could not do had he only some result about $H^1(G,\mu_{p^n}\otimes M)$.

To prove (1), observe that Wiles has managed to make the action of $G$ on $M$ trivial, and $n$ is such that $p^nM=0$. So, $M$ is a finite abelian $p$-group of exponent bounded by $n$, and isomorphic (as $G$-module!) to a finite number of copies of $\mathbb{Z}/p^a\mathbb{Z}$ for $a\leq n$. It thus suffices to prove (1) assuming $M=\mathbb{Z}/p^a$ for $a\leq n$. Consider, for any $s\geq 0$, the exact sequence
$$
0\longrightarrow \mu_{p^s}{\longrightarrow}\mu_{p^{s+a}}\longrightarrow \mu_{p^a}\longrightarrow 0.
$$
By the first result on $H^2$ it gives rise to a surjective map
$$
H^1(G, \mu_{p^s}){\longrightarrow}H^1(G,\mu_{p^{s+a}})\longrightarrow H^1(G,\mu_{p^a})\longrightarrow 0
$$
and, by taking projective limit over $s$,
$$
H^1(G, \mathbb{Z}_p(1))\overset{p^a}{\longrightarrow}H^1(G,\mathbb{Z}_p(1)\longrightarrow H^1(G,\mu_{p^a})\longrightarrow 0
$$
which means $H^1(G, \mathbb{Z}_p(1))/p^aH^1(G, \mathbb{Z}_p(1))\cong H^1(G,\mu_{p^a})$. But this can be rewritten as
$$
H^1(G, \mathbb{Z}_p(1))\otimes\mathbb{Z}/p^a\cong H^1(G, \mathbb{Z}/p^a(1)).
$$
This is (1) for the module $\mathbb{Z}/p^a$ and finished the proof.