Let $B_R\subset\mathbb{R}^n$ denote the $n$-dimensional open ball of radius $R$, centred at the origin. Then we have
$$
\|\partial^\alpha u\|_{L^p(B_R)}
\leq
\max\{1,R^{-\delta-s}\}
\|\langle x\rangle^{\delta+s}\partial^\alpha u\|_{L^p(\mathbb{R}^n)}
\qquad\textrm{for}\quad
|\alpha|=s,
$$
and
$$
\|u\|_{L^p(B_R)} \leq \max\{1,R^{-\delta}\}\|\langle x\rangle^{\delta}u\|_{L^p(\mathbb{R}^n)},
$$
where $\langle x\rangle=\sqrt{1+x^2}$.
Hence, for any $R>0$, the restriction to $B_R$ of any $u\in M^p_{s,\delta}$ is in $W^{s,p}(B_R)$. In other words, the restriction map $J_R:u\mapsto u|_{B_R}$ is well defined and bounded as a map $J_R:M^p_{s,\delta}\to W^{s,p}(B_R)$. By the embedding result for the usual Sobolev spaces, this implies that $M^p_{s,\delta}\subset C(\mathbb{R}^n)$ for $s>\frac np$.
Moreover, by a simple scaling argument it is easy to see that
$$
\|u\|_{L^\infty(A_R)} \leq c \left(R^{-n/p}\|u\|_{L^{p}(A_R)} + R^{s-n/p}|u|_{W^{s,p}(A_R)}\right),
$$
for $s>\frac np$, where $A_R=B_{2R}\setminus \overline{B}_R$ is an annulus. On $A_R$ with $R>1$, we have $\langle x\rangle\sim R$, which implies that
$$
\|u\|_{L^{p}(A_R)} \leq c R^{-\delta}\|u\|_{M^p_{s,\delta}},
\qquad\textrm{and}\qquad
|u|_{W^{s,p}(A_R)} \leq c R^{-\delta-s}\|u\|_{M^p_{s,\delta}},
$$
for $R>1$, and thus
$$
\|u\|_{L^\infty(A_R)} \leq c R^{-\delta-n/p}\|u\|_{M^p_{s,\delta}}.
\qquad\qquad(*)
$$
It shows that if $u\in M^p_{s,\delta}$ with $s>\frac np$ and $\delta>-\frac np$, then $|u(x)|\to0$ uniformly as $x\to\infty$. In particular, we have the continuous embedding $M^p_{s,\delta}\hookrightarrow C_0(\mathbb{R}^n)$.

Let us now discuss the compactness of $M^p_{s,\delta}\hookrightarrow C_0(\mathbb{R}^n)$. By the Rellich-Kondrashov theorem, the restriction map $J_R:M^p_{s,\delta}\to C_b(B_R)$ is compact, with $C_b(B_R)$ denoting the bounded continuous functions on $B_R$ equipped with the uniform norm. Let $\{u_k\}$ be a bounded sequence in $M^p_{s,\delta}$. Then we can extract a subsequence
$$
u_{11},\ldots,u_{1m},\ldots,
$$
such that $\|u_{1m}-u_{1m'}\|_{L^\infty(B_1)}\to0$ as $m,m'\to\infty$. From this subsequence, we can further extract a subsequence
$$
u_{21},\ldots,u_{2m},\ldots,
$$
such that $\|u_{2m}-u_{2m'}\|_{L^\infty(B_2)}\to0$ as $m,m'\to\infty$. We repeat this process to get the doubly-infinite sequence
$$
\begin{split}
u_{11},&\ldots,u_{1m},\ldots,\\
\ldots&\ldots \ldots\\
u_{j1},&\ldots,u_{jm},\ldots,\\
\ldots&\ldots \ldots
\end{split}
\qquad\qquad(**)
$$
which has the property that $\|u_{jm}-u_{jm'}\|_{L^\infty(B_j)}\to0$ as $m,m'\to\infty$.
Now as usual, we look at the diagonal $\{u_{jj}\}$. Let $\varepsilon>0$ be given. Then we choose (an integer) $R$ so large that
$$
\|u_k\|_{L^\infty(\mathbb{R}^n\setminus B_R)}\leq\varepsilon,
$$
for all $k$. This is possible because of the uniform decay $(*)$. Hence
$$
\|u_{jj}-u_{mm}\|_{L^\infty(\mathbb{R}^n)}
\leq
\|u_{jj}-u_{mm}\|_{L^\infty(B_R)}
+
\|u_{jj}-u_{mm}\|_{L^\infty(\mathbb{R}^n\setminus B_R)}
\leq
\|u_{jj}-u_{mm}\|_{L^\infty(B_R)}
+
2\varepsilon.
$$
If $j\geq R$ and $m\geq R$ then both $u_{jj}$ and $u_{mm}$ are elements of the $R$-th row of $(**)$, and since this row is a Cauchy sequence in $L^\infty(B_R)$, we conclude that $\{u_{jj}\}$ is Cauchy in $L^\infty(\mathbb{R}^n)$.