[proofplan]
We construct the test map directly from the join coordinates. A point of the deleted join records weights $\lambda_j$ and points $x_j$ lying in pairwise disjoint faces whenever $\lambda_j>0$; we send the $j$th coordinate to $(\lambda_j,\lambda_j f(x_j)) \in \mathbb{R}^{d+1}$. After subtracting the average vector, the resulting $r$-tuple lies in $W_r^{\oplus(d+1)}$. The hypothesis excluding Tverberg intersections says precisely that this centered vector never vanishes, so normalization gives a map to the sphere, and the formula is symmetric under permutations of the $r$ labels.
[/proofplan]
[step:Define the affine test vector on the deleted join]
Let $Y := K^{*r}_{\Delta}$ be the $r$-fold deleted join. A point $y \in Y$ can be represented in join coordinates as
\begin{align*}
y = \lambda_1 x_1 * \cdots * \lambda_r x_r,
\end{align*}
where $\lambda_j \in [0,1]$, $\sum_{j=1}^{r}\lambda_j = 1$, and, whenever $\lambda_j>0$, the point $x_j$ lies in the geometric realization $|\sigma_j|$ of a face $\sigma_j$ of $K$, with the active faces pairwise disjoint.
Define a map
\begin{align*}
V: Y \to (\mathbb{R}^{d+1})^r
\end{align*}
as follows. For $y=\lambda_1 x_1 * \cdots * \lambda_r x_r$, define
\begin{align*}
v_j(y) := \bigl(\lambda_j,\lambda_j f(x_j)\bigr) \in \mathbb{R}^{d+1}
\end{align*}
for each $j \in \{1,\dots,r\}$, and set
\begin{align*}
V(y) := \bigl(v_1(y),\dots,v_r(y)\bigr).
\end{align*}
If $\lambda_j=0$, then $v_j(y)=(0,0,\dots,0)$, so $v_j(y)$ is independent of the unused representative $x_j$. Therefore $V$ is well-defined on join equivalence classes. On every simplex of $Y$, the functions $\lambda_j$ are affine coordinate functions and $x_j \mapsto f(x_j)$ is continuous, so $v_j$ is continuous on that simplex. The definitions agree on common faces because zero-weight coordinates contribute the zero vector. Hence $V$ is continuous on $Y$.
[/step]
[step:Center the test vector into $W_r^{\oplus(d+1)}$]
Define the averaging map
\begin{align*}
A: (\mathbb{R}^{d+1})^r \to \mathbb{R}^{d+1}
\end{align*}
by
\begin{align*}
A(u_1,\dots,u_r) := \frac{1}{r}\sum_{j=1}^{r}u_j.
\end{align*}
For $y \in Y$, write
\begin{align*}
\bar v(y) := A(V(y)) = \frac{1}{r}\sum_{j=1}^{r}v_j(y).
\end{align*}
Now define
\begin{align*}
\widetilde V: Y \to (\mathbb{R}^{d+1})^r
\end{align*}
by
\begin{align*}
\widetilde V(y) := \bigl(v_1(y)-\bar v(y),\dots,v_r(y)-\bar v(y)\bigr).
\end{align*}
The map $\widetilde V$ is continuous because it is obtained from the continuous map $V$ by addition and scalar multiplication in finite-dimensional Euclidean space. Also,
\begin{align*}
\sum_{j=1}^{r}\bigl(v_j(y)-\bar v(y)\bigr) = \sum_{j=1}^{r}v_j(y)-r\bar v(y)=0.
\end{align*}
Thus $\widetilde V(y)$ lies in the subspace
\begin{align*}
\left\{(u_1,\dots,u_r)\in(\mathbb{R}^{d+1})^r:\sum_{j=1}^{r}u_j=0\right\},
\end{align*}
which is the standard identification of $W_r^{\oplus(d+1)}$ with centered $r$-tuples in $\mathbb{R}^{d+1}$.
[/step]
[step:Use the no intersection hypothesis to prove the centered vector is nonzero]
We claim that $\widetilde V(y)\neq 0$ for every $y\in Y$. Suppose, toward a contradiction, that $\widetilde V(y)=0$ for some
\begin{align*}
y=\lambda_1x_1*\cdots *\lambda_rx_r \in Y.
\end{align*}
Then $v_1(y)=\cdots=v_r(y)=\bar v(y)$. Comparing first coordinates gives
\begin{align*}
\lambda_1=\lambda_2=\cdots=\lambda_r.
\end{align*}
Since $\sum_{j=1}^{r}\lambda_j=1$, each weight equals $1/r$. In particular, all weights are positive, so each $x_j$ lies in an active face $|\sigma_j|$, and the faces $\sigma_1,\dots,\sigma_r$ are pairwise disjoint by the definition of the deleted join.
Comparing the last $d$ coordinates of the equalities $v_1(y)=\cdots=v_r(y)$ gives
\begin{align*}
\lambda_1 f(x_1)=\lambda_2 f(x_2)=\cdots=\lambda_r f(x_r).
\end{align*}
Since $\lambda_j=1/r$ for every $j$, multiplying by $r$ yields
\begin{align*}
f(x_1)=f(x_2)=\cdots=f(x_r).
\end{align*}
Thus this common point belongs to $f(|\sigma_1|)\cap\cdots\cap f(|\sigma_r|)$, contradicting the hypothesis. Therefore $\widetilde V(y)\neq 0$ for all $y\in Y$.
[guided]
The only possible obstruction to normalizing $\widetilde V$ is that it might vanish somewhere. We now show that a zero of $\widetilde V$ is exactly the same kind of configuration forbidden by the theorem hypothesis.
Assume that $\widetilde V(y)=0$ for some point
\begin{align*}
y=\lambda_1x_1*\cdots *\lambda_rx_r \in K^{*r}_{\Delta}.
\end{align*}
By definition,
\begin{align*}
\widetilde V(y)=\bigl(v_1(y)-\bar v(y),\dots,v_r(y)-\bar v(y)\bigr),
\end{align*}
so $\widetilde V(y)=0$ means that every component satisfies $v_j(y)=\bar v(y)$. Hence
\begin{align*}
v_1(y)=v_2(y)=\cdots=v_r(y).
\end{align*}
Each vector $v_j(y)$ has first coordinate $\lambda_j$, because
\begin{align*}
v_j(y)=\bigl(\lambda_j,\lambda_j f(x_j)\bigr).
\end{align*}
Therefore equality of the vectors $v_j(y)$ implies equality of their first coordinates:
\begin{align*}
\lambda_1=\lambda_2=\cdots=\lambda_r.
\end{align*}
The join weights also satisfy $\sum_{j=1}^{r}\lambda_j=1$, so the common value is $1/r$. Thus every $\lambda_j$ is positive. This matters because the deleted join condition only controls the coordinates with positive weights: since all weights are positive, every $x_j$ is an active point lying in a face $|\sigma_j|$, and the faces $\sigma_1,\dots,\sigma_r$ are pairwise disjoint.
Now compare the remaining $d$ coordinates of the same vector equality. Since the last $d$ coordinates of $v_j(y)$ are $\lambda_j f(x_j)$, we get
\begin{align*}
\lambda_1 f(x_1)=\lambda_2 f(x_2)=\cdots=\lambda_r f(x_r).
\end{align*}
Using $\lambda_j=1/r$ for all $j$, we multiply by $r$ and obtain
\begin{align*}
f(x_1)=f(x_2)=\cdots=f(x_r).
\end{align*}
The common value therefore lies in every image $f(|\sigma_j|)$. Hence
\begin{align*}
f(|\sigma_1|)\cap\cdots\cap f(|\sigma_r|)\neq\varnothing.
\end{align*}
This contradicts the assumed absence of common intersections among images of $r$ pairwise disjoint faces. Consequently $\widetilde V$ has no zero on $K^{*r}_{\Delta}$.
[/guided]
[/step]
[step:Normalize to obtain the equivariant sphere map]
Define
\begin{align*}
\Phi: Y \to S(W_r^{\oplus(d+1)})
\end{align*}
by
\begin{align*}
\Phi(y) := \frac{\widetilde V(y)}{|\widetilde V(y)|}.
\end{align*}
The denominator is nonzero by the previous step, so $\Phi$ is well-defined. Since $\widetilde V$ is continuous and the Euclidean norm is continuous and positive on the image of $\widetilde V$, the map $\Phi$ is continuous. Its image lies in the unit sphere because
\begin{align*}
|\Phi(y)|=1
\end{align*}
for every $y\in Y$.
[/step]
[step:Check that the construction commutes with the $S_r$ action]
Let $S_r$ act on $Y=K^{*r}_{\Delta}$ by permuting the $r$ join factors, and let it act on $W_r^{\oplus(d+1)}$ by permuting the $r$ vector components. For $\pi\in S_r$ and $y=\lambda_1x_1*\cdots *\lambda_rx_r$, the point $\pi y$ has $j$th join coordinate equal to the original $\pi^{-1}(j)$th coordinate. Therefore
\begin{align*}
v_j(\pi y)=v_{\pi^{-1}(j)}(y).
\end{align*}
The average is permutation-invariant, since
\begin{align*}
\bar v(\pi y)=\frac{1}{r}\sum_{j=1}^{r}v_j(\pi y)=\frac{1}{r}\sum_{j=1}^{r}v_j(y)=\bar v(y).
\end{align*}
Hence
\begin{align*}
\widetilde V(\pi y)=\pi \widetilde V(y).
\end{align*}
The Euclidean norm is invariant under coordinate permutations, so
\begin{align*}
\Phi(\pi y)=\frac{\widetilde V(\pi y)}{|\widetilde V(\pi y)|}=\frac{\pi\widetilde V(y)}{|\widetilde V(y)|}=\pi\Phi(y).
\end{align*}
Thus $\Phi$ is $S_r$-equivariant. This proves the existence of the required $S_r$-equivariant continuous map
\begin{align*}
K^{*r}_{\Delta} \to S(W_r^{\oplus(d+1)}).
\end{align*}
[/step]
