I will here spell out more explicitly what the vertices of the local polytope look like.
The local polytope is generated by local deterministic behaviours
The local polytope is, by definition, comprised of behaviours $p(ab|xy)$ such that
$$p(ab|xy)=\sum_\lambda p_\lambda p_\lambda(a|x) p_\lambda(b|y).$$
Moreover, for each $\lambda$ and $x$, we can always decompose $p_\lambda(a|x)$ as a convex combination of local deterministic assignments. For example,
if $p_\lambda(0|x)=q_{\lambda,0x}$ and $p_\lambda(1|x)=1-q_{\lambda,0x}$, then we can write
$$p_\lambda(a|x) = q_{\lambda,0x} \delta_{a,0} + (1-q_{\lambda,0x})\delta_{a,1}.$$
As a vector, this reads
\begin{align}\mathbf p_\lambda &=
(q_{\lambda,00}\mathbf e_{0} + (1-q_{\lambda,00})\mathbf e_{1})
\oplus
(q_{\lambda,01}\mathbf e_{0} + (1-q_{\lambda,01})\mathbf e_{1}) \\
&= (q_{\lambda,00},1-q_{\lambda,00},q_{\lambda,01},1-q_{\lambda,01}) \\
&= (p_\lambda(0|0), p_\lambda(1|0), p_\lambda(0|1), p_\lambda(1|1)),
\end{align}
for $q_{\lambda,00},q_{\lambda,01}\in\{0,1\}$.
All such vectors are generated by the basis vectors
$\mathbf e_{ij}\equiv \mathbf e_i\oplus\mathbf e_j$, representing behaviours in which $x=0$ corresponds to $a=i$ and $x=1$ to $a=j$. Explicitly, these are spanned by the four length-four vectors:
$$(1,0,1,0),\quad (1,0,0,1),\quad (0,1,1,0), \quad (0,1,0,1).$$
(Behaviours in full space) Now, what about the actual behaviours corresponding to the full conditional probabilities $p(ab|xy)$? To accommodate any such distribution, these must be vectors with $(2^2)^{2^2}=2^8=256$ elements. The deterministic behaviours in such space are the vectors of the form
$$\underbrace{\mathbf e_{ij}}_{x=0,y=0}\otimes \underbrace{\mathbf e_{k\ell}}_{x=0,y=1}\otimes \underbrace{\mathbf e_{mn}}_{x=1,y=0}\otimes\underbrace{\mathbf e_{pq}}_{x=1,y=1},$$
with each $\mathbf e_{ij}$ characterising the outputs $ab$ associated with a given pair of inputs $xy$.
Clearly, however, such vectors do not always correspond to local behaviours. For example, the vector
$$\mathbf e_{00}\otimes \mathbf e_{10}\otimes \mathbf e_{\bullet\bullet}\otimes \mathbf e_{\bullet\bullet}$$
is not local, as it gives $(a,b)=(0,0)$ for $(x,y)=(0,0)$ but $(a,b)=(1,0)$ for $(x,y)=(0,1)$.
(Local behaviours in full space)
Local behaviours have the form
$$\mathbf e_{a_0b_0}\otimes
\mathbf e_{a_0b_1} \otimes
\mathbf e_{a_1b_0} \otimes
\mathbf e_{a_1b_1} \simeq \mathbf e_{a_0 a_1}\otimes\mathbf e_{b_0b_1},
$$
for some choice of $a_i,b_j\in\{0,1\}$. There are therefore only $(2^2)^2=16$ such basic vectors, whose convex combinations draw the subset of local behaviours in the bigger $256$-dimensional space of general behaviours.
Show that all local deterministic behaviours are vertices
Suppose
$$\left|\sum_{abxy} (-1)^{a+b+xy} P(ab|xy)\right|= 2.$$
Explicitly, this amounts to the following two equations (I'll use $P_{ab,xy}\equiv P(ab|xy)$ for notational brevity):
$$
(P_{00,00} + P_{11,00} - P_{01,00} - P_{10,00}) +
(P_{00,01} + P_{11,01} - P_{01,01} - P_{10,01}) +
(P_{00,10} + P_{11,10} - P_{01,10} - P_{10,10}) -
(P_{00,11} + P_{11,11} - P_{01,11} - P_{10,11}) = \pm2.
$$
Taking into account the normalisation condition for each $(x,y)$, we can simplify this to
$$
(\underbrace{P_{00,00} + P_{11,00}}_{\equiv a_{00}}) +
(\underbrace{P_{00,01} + P_{11,01}}_{\equiv a_{01}}) +
(\underbrace{P_{00,10} + P_{11,10}}_{\equiv a_{10}}) -
(\underbrace{P_{00,11} + P_{11,11}}_{\equiv a_{11}}) \in \{0,2\}.
\tag X$$
Consider the possible deterministic assignments. Any such assignment corresponds to a choice of each of the above four terms on the LHS $-$ here denoted with $a_{xy}$ $-$ equaling one (if $P_{00,xy}=1$ or $P_{11,xy}=1$) or zero (if $P_{01,xy}=1$ or $P_{10,xy}=1$).
In other words, $a_{xy}\in\{0,1\}$ for all $x,y$.
The possible combinations are thus seen to be
$$
\begin{array}{c|c|c|c|c}
a_{00} & a_{01} & a_{10} & a_{11} & S\\\hline
1 & 0 & 0 & 1 & 0\\\hline
0 & 1 & 0 & 1 & 0 \\\hline
0 & 0 & 1 & 1 & 0 \\\hline
1 & 1 & 1 & 1 & 2 \\\hline
0 & 0 & 0 & 0 & 0 \\\hline
1 & 1 & 0 & 0 & 2 \\\hline
1 & 0 & 1 & 0 & 2 \\\hline
0 & 1 & 1 & 0 & 2 \\\hline
\end{array}
$$
Each of these rows corresponds to $2^4=16$ deterministic behaviours. However, this does not take into account the locality constraint.
To see this, focus for example on the first row. A compatible behaviour corresponding to this row is:
$$\mathbf e_{00}\otimes\mathbf e_{01}\otimes\mathbf e_{10}\otimes\mathbf e_{00}.$$
This is nonlocal: the fourth assignment shouldn't be possible, as the first three established that $a=x$ and $b=y$, but here we get that $(1,1)\to(0,0)$.
More generally, a local deterministic behaviour must be factorisable: $$P(ab|xy)=p(a|x)q(b|y)\equiv p_{a|x}q_{b|y}=\delta_{a,a_x}\delta_{b,b_y},$$
for some deterministic probability distributions $p$ and $q$. There are $(2^2)^2=16$ such assignments,
and the proper way to list them is via each possible assignment of four bits to the four variables $(a_0, a_1, b_0, b_1)$.
Let us rewrite (X) embedding the locality constraint in the expression. We get
$$
(p_{00} + p_{01}) q_{00} +
(p_{00} - p_{01}) q_{01} +
(p_{10} + p_{11}) q_{10} +
(p_{10} - p_{11}) q_{11} \in \{0,2\}.
$$
One can verify directly that each one of the possible $16$ assignments satisfies this equation, meaning that all possible deterministic local behaviours are vertices on the local polytope.
Half of these are vertices for the polytope corresponding to $S=2$, and the remaining $8$ for the $S=0$ polytope.
