The cylindroid C need not be a manifold with corners. Here's a sketch of a counterexample (different from the incorrect one I posted earlier). Forgive my crude drawings.
Begin with a surface in $\mathbb R^3$ that's diffeomorphic to a disk. Looking down on it from the positive $z$-axis, its projection on the $xy$-plane looks like this:
But in 3-d, it actually lies on the surface of a helicoid, so as you walk counterclockwise around the surface, you gradually rise one unit in the $z$-direction. The darkened point on the positive $x$-axis actually represents two points on the surface, with one of them $1$ unit above the other.
Now let $\gamma:[0,1]\to \mathbb R^3$ be the curve $\gamma(t) = (0,0,t)$. The cylindroid $C$ is thus obtained by taking this helicoidal surface and translating it directly upward one unit. The resulting solid figure looks roughly like this:
The red dot is the point where the original surface meets its translated version. There is no neighborhood of the red dot in $C$ that is homeomorphic to an open subset of $\mathbb R^3$ or of $H^3_k$ for any $k$.
ADDED: Assume for contradiction that $C$ is a manifold with corners. Since some points of $C$ have neighborhoods homeomorphic to open balls in $\mathbb R^3$, it must be $3$-dimensional by invariance of dimension. Our assumption implies that there is a (relative) neighborhood $U$ of the red point in $C$ and a homeomorphism $\varphi\colon U\to \widehat U$, where $\widehat U$ is an open subset of $\mathbb R^3$ or of $H^3_k$ for some $k$. By shrinking $U$ and $\widehat U$ if necessary, we can assume both are connected. Shrinking further, we can assume that $U$ is contained in the ball of radius $1/8$, say, around the red point.
Now let $q = \varphi(\text{red point})$. Then $\widehat U\smallsetminus\{q\}$ is connected, but $U\smallsetminus \{\text{red point}\}$ is disconnected. Since homeomorphisms preserve connectivity, this is a contradiction.
Above we showed that the subset $\overset{\circ}{\mathcal E}_k$ of $\mathcal E_k$ whose elements have value in $[0,1)$ with respect the scalar function $g$ is open in $H^k_k$ since it is union of the sets
$$
g^{-1}\big[\,(0,1)\,\big]\cap H^k_k\,\,\,\text{and}\,\,\,C\Big(O,\frac 1k\Big)\cap H^k_k
$$
that are open in $H^k_k$. Now provided that
$$
\xi^1+\dots+\xi^k=1
$$
for any $\xi\in\mathcal E_k$ then surely it must be
$$
\xi^j\gneq0
$$
for any $j=1,\dots,k$ and in particular for now we assume that this happens for the last coordiante. So in this case we define the map $\varphi$ from $\Bbb R^k$ into $\Bbb R^k$ through the equation
$$
\varphi(x):=\big(x^1,\dots,x^{k-1},1-(x^1+\dots+x^{k-1}+x^k)\big)
$$
for any $x\in\Bbb R^k$ and thus we are going to prove that this map can be restricted to a coordinate patch about $\xi\in\mathcal E_k$ whose last coordinate is not zero, i.e. we let to prove that the restriction $\phi$ of $\varphi$ at the set $\overset{\circ}{\mathcal E}_k$ is the researched coordinate patch. First of all we observe that $\varphi$ is a diffeomorphism of $\Bbb R^k$ onto $\Bbb R^k$ being an affine map, i.e. it is compostion of a translation with a linear map between finite dimensional topological vector spaces and both these maps are diffeomorphism. So if we prove that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ into an open set of $\mathcal E_k$ containing $\xi$ then we will proved that $\varphi$ is a coordinate patch about $\xi$ so that we let to do this. Now if $x$ is an elemen of $\overset{\circ}{\mathcal F}$ then it must be
$$
x^i\in[0,1)
$$
for each $i=1,\dots,k$ since otherwise it would be
$$
x^1+\dots+x^{i-1}+1+x^{i+1}+\dots+x^k\le x^1+\dots+x^{i-1}+x^i+x^{i+1}+\dots+x^k<1\Rightarrow\\ x^1+\dots+x^{i-1}+x^{i+1}+\dots+x^k<0
$$
that is impossible if $x$ lies in $\overset{\circ}{\mathcal E_k}$ and thus in $H^k_k$ and so we can conclude that
$$
x\in\overset{\circ}{\mathcal E}_k\Rightarrow x\in[0,1)^k\wedge x^1+\dots+x^k\lneq1\Rightarrow\\\varphi(x)\in H^k_k\wedge\varphi^1(x)+\dots+\varphi^k(x)\le1\Rightarrow\varphi(x)\in\mathcal E_k
$$
and this proves that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ into $\mathcal E_k$. So if the last coordinate of $\xi$ is not zero and the others are not negative then surely
$$
g(\xi^1,\dots,\xi^{k-1},0)\ge0\,\,\,\text{and}\,\,\,g(\xi^1,\dots,\xi^{k-1},0)=\xi^1+\dots+\xi^{k-1}<\xi^1+\dots+\xi^{k-1}+\xi^k=1
$$
so that $(\xi^1,\dots,\xi^{k-1},0)$ is an element of $\overset{\circ}{\mathcal E}_k$ and in particular it is such that
$$
\varphi(\xi^1,\dots,\xi^{k-1},0)=\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1})\big)=\big(\xi^1,\dots,\xi^{k-1},\xi^k\big)=\xi
$$
and thus it is proved that $\xi$ is an element of $\varphi\big[\,\overset{\circ}{\mathcal E}_k\,\big]$. Now we previously define the set
$$
\tilde{\mathcal E}_k:=\{x\in\mathcal E_k:x^k\neq 0\}
$$
and thus we let prove that $\varphi$ carries $\overset{\circ}{\mathcal E}_k$ onto $\tilde{\mathcal E}_k$. So we first observe that
$$
x\in\overset{\circ}{\mathcal E}_k\Rightarrow 0\le g(x)\lneq1\Rightarrow0\lneq1-g(x)\le1\Rightarrow\varphi^k(x)>0\Rightarrow\varphi(x)\in\tilde{\mathcal E}_k
$$
and so we conclude that
$$
\varphi\big[\,\overset{\circ}{\mathcal E}_k\,\big]\subseteq\tilde{\mathcal E}_k
$$
so that we let to prove the opposite inclusion. Now for any element $\xi$ of $\tilde{\mathcal E}_k$ it must be
$$
g(\xi)=1\,\,\,\text{or either}\,\,\,g(\xi)<1
$$
but effectively we just proved above that if $g(\xi)=1$ and $\xi^k>0$ then surely
$$
\xi=\varphi(x)
$$
for any $x\in\overset{\circ}{\mathcal E}$ so that to follow we are supposing that $g(\xi)$ is strictly less than $1$. So we observe that
$$
\begin{cases}\xi\in\tilde{\mathcal E}_k\\
g(\xi)\lneq1\end{cases}\Rightarrow
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\\xi^k\gneq0\\
\xi^1+\dots+\xi^{k-1}+\xi^k\lneq1\end{cases}\Rightarrow\\
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\0\lneq\xi^k\lneq1\\
0\lneq1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\lneq1\end{cases}\Rightarrow
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\0\lneq1-\xi^k\lneq1\\
0\lneq1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\lneq1\end{cases}\Rightarrow
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\
1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\gneq0\\
0\lneq\xi^1+\dots+\xi^{k-1}+\big(1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)\lneq1\end{cases}
$$
and thus we conclude that $\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)$ is an element of $\overset{\circ}{\mathcal E}_k$ and so observing that
$$
\varphi\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\biggl)=\\
\biggl(\xi^1,\dots,\xi^{k-1},1-\Big(\xi^1+\dots+\xi^{k-1}+\big(1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)\Big)\biggl)=\\
\Biggl(\xi^1,\dots,\xi^{k-1},1-\Big((\xi^1+\dots+\xi^{k-1})+1-(\xi^1+\dots+\xi^{k-1})-\xi^k\Big)\Biggl)=\\
\big(\xi^1,\dots,\xi^{k-1},1-(1-\xi^k)\big)=(\xi^1,\dots,\xi^{k-1},\xi^k)=\xi$$
we conclude that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ onto $\tilde{\mathcal E}_k$. So we let to prove that $\tilde{\mathcal E}_k$ is open in $\mathcal E_k$ and we are doing this proving that the set $\mathcal E_k\setminus\tilde{\mathcal E}_k$ is closed in $\mathcal E_k$, that is it is the intersection of a closed set of $\Bbb R^k$ with $\mathcal E_k$. Now we remember that any symplex is closed (see here for details) so that the $(k-1)$ standard simplex $\mathcal E_{k-1}$ is closed: so observing that
$$
\mathcal E_k\setminus\tilde{\mathcal E}_k=\{x\in\mathcal E_k:x^k=0\}=\\
\{x\in\Bbb R^k:x^1+\dots x^{k-1}\le1\wedge x^i\ge0,\,\forall\,i=1,\dots,(k-1)\wedge x^k=0\}=\\
\{x\in\Bbb R^{k-1}:x^1+\dots+x^{k-1}\le1\wedge x^i\,\forall\,i=1,\dots,(k-1)\ge0\}\times\{0\}=\mathcal E_{k-1}\times\{0\}
$$
we conclude that $\mathcal E_k\setminus\tilde{\mathcal E}_k$ is closed in $\Bbb R^k$ (ideed it is product of closed set) and so $\mathcal E_k$ too. So we proved the existence of a coordinate patch when the last coordinate of $\xi$ is not zero. Otherwise if the last coordinate of $\xi$ is zero then surely (remember what observed above) there must exist $j=1,\dots,k-1$ such that
$$
\xi^j\neq0
$$
so that let be $\psi$ the diffeormosphism of $\Bbb R^k$ onto $\Bbb R^k$ that interchanges the $j$-th coordinate with the last, i.e.
$$
[\psi(x)](i):=\begin{cases}x^k,\,\,\,\text{if}\,\,\,i=j\\
x^j,\,\,\,\text{if}\,\,\,i=k\\
x^i\,\,\,\text{otherwise}\end{cases}
$$
for any $x\in\Bbb R^k$. Now $\psi$ is an involution that maps $\mathcal E_k$ onto $\mathcal E_k$ and in particular if the last coordinate of $\xi$ is zero then this does not happen for $\psi^{-1}(\xi)$ so that if $\varphi$ is a coordinate patch about $\psi^{-1}(\xi)$ then the composition $\psi\circ\varphi$ of $\psi$ with $\varphi$ is just a coordinate patch about $\xi$ and so the statement finally holds.
Best Answer
So in the posted link it is shown that any coordinate patch of $\mathcal S$ is equal to the composition of an affine map $f:\Bbb R^k\rightarrow\Bbb R^n$ with a coordinate patch $\alpha:U\rightarrow V$ of $\mathcal E_k$ and thus if $(f\circ\alpha_1)$ and $(f\circ\alpha_2)$ are two coordinate patch that overlap then $$ (f\circ\alpha_2)^{-1}\circ(f\circ\alpha_1)=\alpha_2^{-1}\circ f^{-1}\circ f\circ\alpha_1=\alpha_2^{-1}\circ\alpha_1 $$ so that if we prove that $\mathcal E_k$ is orientable the even $\mathcal S$ will be orientable but $\mathcal E_k$ is trivially orientable because it is a $k$-manifold in $\Bbb R^k$ and any such manifold can be equipped with the natural orientation as here showed.