If you study the proof of Noether's normalization lemma, you'll find that it's actually a constructive proof - it provides an algorithm to solve your problem. I'm going to do it for you; see if you can figure out what's going on. (I'm not going to describe the algorithm in detail, but if you pay attention you should be able to understand why I'm doing what I'm doing.)
First, note that we should expect $m=2$; indeed, the variety defined by a single irreducible polynomial has codimension 1. So it will be a "surface" in $k^3$. We should therefore expect the coordinate ring to have transcendence degree $2$ over $k$.
Consider $f$ as a polynomial in $x_3$. The leading coefficient is $x_1+x_2$. Find a point where it doesn't vanish identically, say $x_1=1,x_2=0$. Then put $u_1=x_1+x_3$, $u_2=x_2$. Then we can rewrite $f$ as
$$f=x_3(u_1-x_3+u_2)+(u_1-x_3)u_2 = -x_3^2+x_3u_1 + u_1u_2.$$
Magic! This is a monic polynomial in $x_3$, with coefficients in $k[u_1,u_2]$. Thus $R=k[x_1,x_2,x_3]/(f)$ is integral (of degree $2$) over $k[x_1+x_3,x_2]$.
Geometrically, this corresponds to projecting the surface $f=0$ on the plane spanned by the vectors $(1,0,1)$ and $(0,1,0)$ (i.e. the plane $x-z=0$):
As you can see, the projection is surjective (a consequence of the integrality of the extension), and the fibers have two points almost everywhere.
Let $X_0, X_1, \dots, X_n$ be i.i.d. standard normal vectors in $\mathbb{R}^n$ (so each $X_i \sim \mathcal{N}(0, I_n)$). Writing $Y_i = X_i - X_0$ for $i = 1, \dots, n$, we have that the $n$-volume of the $n$-simplex with vertices $X_0, X_1, \dots, X_n$ is equal to
$$\frac{1}{n!} |\det(Y_1, \dots, Y_n)|$$
where we consider $Y_1, \dots, Y_n$ as column vectors.
Define $(W_1, \dots, W_n) = (Y_1, \dots, Y_n)^T$, i.e. $W_{i, j} = X_{j, i} - X_{0, i}$, so $W_1, \dots, W_n$ are independent, and $W_i \sim \mathcal{N}(0, \Sigma)$, where the covariance matrix $\Sigma$ has $2$'s on the diagonal and $1$'s off the diagonal. Note that $J_n$ (the matrix of ones) has eigenvalues $n, 0, \dots, 0$, hence since $\Sigma = I_n + J_n$, $\Sigma$ has eigenvalues $n+1, 1, \dots, 1$ and thus $\det \Sigma = n+1$. Now, defining $Z_i = \Sigma^{-1/2} W_i$ for $i = 1, \dots, n$, we have that $Z_1, \dots, Z_n$ are independent with each $Z_i \sim \mathcal{N}(0, I_n)$, and also that
$$\det(Y_1, \dots, Y_n) = \det(W_1, \dots, W_n) = \det(\Sigma^{1/2}Z_1, \dots, \Sigma^{1/2}Z_n) = \det \Sigma^{1/2} \cdot \det(Z_1, \dots, Z_n).$$
It follows that the desired expected volume is
$$\frac{\sqrt{n+1}}{n!} \mathbb{E}[|\det(Z_1, \dots, Z_n)|]$$
for independent $Z_1, \dots, Z_n \sim \mathcal{N}(0, I_n)$. To finish, we compute $\mathbb{E}[|\det(Z_1, \dots, Z_n)|]$.
Let $Z_1', \dots, Z_n'$ be the result of performing the Gram-Schmidt process to $Z_1, \dots, Z_n$ without normalizing, so for each $k$, we have $\mathrm{span}(Z_1', \dots, Z_k') = \mathrm{span}(Z_1, \dots, Z_k)$, and we inductively define $Z_k' = Z_k - P_kZ_k$ (with $Z_1' = Z_1$), where $P_k$ is the orthogonal projection onto $\mathrm{span}(Z_1', \dots, Z_{k-1}')$. Notably, these are all elementary column operations, so $\det(Z_1', \dots, Z_n') = \det(Z_1, \dots, Z_n)$, and $Z_1', \dots, Z_n'$ are orthogonal, so $|\det(Z_1', \dots, Z_n')| = \prod_{k=1}^n |Z_k'|$. Equivalently, we have $Z_k' = P_k' Z_k$, where $P_k'$ is the orthogonal projection onto the orthogonal complement of $\mathrm{span}(Z_1', \dots, Z_{k-1}')$, so $Z_k'$ can be seen as a standard normal vector on this $(n-k+1)$-dimensional space. This means that conditioning on $Z_1', \dots, Z_{k-1}'$, $|Z_k'|$ has the chi distribution with $n-k+1$ degrees of freedom, so in fact $|Z_k'|$ is independent of $Z_1', \dots, Z_{k-1}'$ with
$$\mathbb{E}[|Z_k'|] = \sqrt{2} \frac{\Gamma((n-k+2)/2)}{\Gamma((n-k+1)/2)}.$$
It follows that all $|Z_k'|$ are independent, giving
\begin{align*}
\mathbb{E}[|\det(Z_1, \dots, Z_n)|]
&= \prod_{k=1}^n \mathbb{E}[|Z_k'|]\\
&= \prod_{k=1}^n \sqrt{2} \frac{\Gamma((n-k+2)/2)}{\Gamma((n-k+1)/2)} \\
&= \prod_{k=1}^n \sqrt{2} \frac{\Gamma((k+1)/2)}{\Gamma(k/2)} \\
&= 2^{n/2} \frac{\Gamma((n+1)/2)}{\Gamma(1/2)}
\end{align*}
so the expected volume is $2^{n/2} \frac{\Gamma((n+1)/2) \sqrt{n+1} }{\Gamma(1/2) n!}$. At $n = 3$ (the given case), this is $\frac{2}{3} \sqrt{\frac{2}{\pi}}$.
Higher moments can be computed in the same way, using the corresponding higher moments of the chi distribution.
Best Answer
Visualization: The hyperplane $H_m^n$ is parallel to one facet of the simplex, and a simplex is a cone; their intersection is a scaled-down copy of the facet. The vertex of the intersection corresponding to $e_1$ is closer to $e_1$ than the rest of the vertices, so a sphere of suitable radius centred at $e_1$ will have one vertex inside and the rest outside, and so will meet the intersection at more than one point.
Computation: Let $e_i$ denote the standard basis vectors. The set $\triangle^{n-1}\cap H_m^n$ is convex and contains the points $$ \{ me_i + (1-m)e_n : i=1,\dotsc,n-1 \} $$ (In fact it is exactly the convex hull of these points, but we won't need that fact.) Let $v_i = me_i + (1-m)e_n$. We have $$ \|v_i-e_1\| = \begin{cases} \sqrt2(1-m) &\text{if $i=1$,} \\ \sqrt2\sqrt{(1-m)^2+m} &\text{if $i\ne 1$.} \end{cases} $$ So if we take $k\in[0,1]$ such that $$ \sqrt2(1-m) < k < \sqrt2\sqrt{(1-m)^2+m} $$ (and $m\in[1-\frac1{\sqrt2},1]$ so that this is possible), then by continuity the sphere will meet $\triangle^{n-1}\cap H_m^n$ at points of the form $(1-\lambda_i)v_1+\lambda_iv_i$ for each $i=2,\dotsc,n-1$, for suitable $\lambda_i\in[0,1]$.
Unpacking that last part a bit as requested: Define $f\colon[0,1]\to\mathbb R^n$ by $f(\lambda) = \|(1-\lambda)v_1+\lambda v_i-e_1\|$. Then $$ f(0) = \sqrt2(1-m) < k < \sqrt2\sqrt{(1-m)^2+m} = f(1) $$ By the intermediate value theorem, there exists $\lambda_i\in[0,1]$ such that $f(\lambda_i)=k$, which means $(1-\lambda_i)v_1+\lambda_i v_i\in S_k^n$.