Wikipedia's definition of the Schwartz class is a bit awkward and I think that is causing some of the difficulty. They define $f \in \mathcal{S}(\mathbb{R}^n)$ if $$\sup_{x \in \mathbb{R}^n} |x^\beta D^\alpha f| < \infty \label{1}\tag{*}$$ for all multi-indices $\alpha, \beta$. A more convenient equivalent definition, found in e.g. Folland's *Real Analysis*, is to have $$\sup_{x \in \mathbb{R}^n} |(1+|x|)^k D^\alpha f| < \infty \label{2}\tag{**}$$ for all multi-indices $\alpha$ and positive integers $k$.

With definition \eqref{2} your (1) becomes easy, writing
$$|r^k g(r)| \le \int_{S^{n-1}} r^k |f(rw)|\,d\mu(w)$$
and noting
$$r^k |f(rw)| \le (1+r)^k |f(rw)| = \left|(1+|rw|)^k f(rw)\right|$$
which by assumption is bounded. You should also be able to get (2) by differentiating under the integral sign and noting that $\frac{d^m}{dr^m} f(rw)$ can be written in terms of the partial derivatives of $f$ up to order $m$, with coefficients involving the coordinates of $w$ which are all bounded by 1.

To see \eqref{1} and \eqref{2} are equivalent, it suffices to take $\alpha=0$. Suppose \eqref{1} holds. It suffices to prove \eqref{2} for even $k$, so replace $k$ by $2k$. Since $1+|x| \le C (1+|x|^2)^{1/2}$ for some universal $C$, it is enough to show $\sup_x |(1+|x|^2)^k f| < \infty$. But $(1+|x|^2)^k$ is a polynomial in $x$ of degree $2k$, so it can be written as a linear combination of monomials $x^\beta$ which can be controlled by \eqref{1}.

Conversely, if \eqref{2} holds, note that $|x_i^{\beta_i}| = |x_i|^{\beta_i} \le (1+|x|)^{\beta_i}$, since $|x_i| \le |x| \le 1+|x|$. So $$|x^\beta| = |x_1^{\beta_1} \dots x_n^{\beta_n}| \le (1+|x|)^{\beta_1 + \dots + \beta_n}$$
which can be controlled by \eqref{1}.

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