= Solution
Write an element of the diagonal torus as
$$
\operatorname{diag}(t_1,\ldots,t_n,-t_n,\ldots,-t_1)
$$
and let $\varepsilon_i$ extract $t_i$. The <root-space decomposition> is
$$
\mathfrak{so}_{2n}
=\mathfrak t\oplus
\bigoplus_{1\leq i<j\leq n}
\left(
\mathfrak g_{\varepsilon_i-\varepsilon_j}\oplus
\mathfrak g_{\varepsilon_i+\varepsilon_j}\oplus
\mathfrak g_{-\varepsilon_i+\varepsilon_j}\oplus
\mathfrak g_{-\varepsilon_i-\varepsilon_j}
\right),
$$
with one-dimensional root spaces. Thus the $D_n$ <root system> is
$$
R=\{\pm\varepsilon_i\pm\varepsilon_j:i<j\}.
$$
The upper-triangular choice gives
$$
R^+=\{\varepsilon_i-\varepsilon_j,\ \varepsilon_i+\varepsilon_j:i<j\}.
$$
A compatible simple system is
$$
\alpha_i=\varepsilon_i-\varepsilon_{i+1}\quad(1\leq i<n),\qquad
\alpha_n=\varepsilon_{n-1}+\varepsilon_n.
$$
The <highest root> and <Weyl vector> are
$$
\theta=\varepsilon_1+\varepsilon_2
=\alpha_1+2\alpha_2+\cdots+2\alpha_{n-2}+\alpha_{n-1}+\alpha_n,
\qquad
\rho=\sum_{i=1}^n(n-i)\varepsilon_i.
$$
The <fundamental weight>[fundamental weights] are
$$
\omega_k=\varepsilon_1+\cdots+\varepsilon_k\quad(1\leq k\leq n-2),
$$
$$
\omega_{n-1}=\frac12(\varepsilon_1+\cdots+\varepsilon_{n-1}-\varepsilon_n),
\qquad
\omega_n=\frac12(\varepsilon_1+\cdots+\varepsilon_n).
$$
Using the paper's letters, the root lattice and weight lattice are respectively
$$
P=\left\{(a_i)\in\mathbb Z^n:\sum_i a_i\ \text{is even}\right\},
\qquad
Q=\mathbb Z^n\cup\left(\mathbb Z+\frac12\right)^n.
$$
Their quotient is
$$
Q/P\cong
\begin{cases}
\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z,&n\text{ even},\\
\mathbb Z/4\mathbb Z,&n\text{ odd}.
\end{cases}
$$
The <Dynkin diagram> is the $D_n$ diagram: a chain $\alpha_1-\cdots-\alpha_{n-2}$ whose last node is joined to both $\alpha_{n-1}$ and $\alpha_n$. The <Extended Dynkin diagram> adds $\alpha_0=-\theta$ joined to $\alpha_2$. For $D_4$, the central node $\alpha_2$ consequently has the four leaves $\alpha_0,\alpha_1,\alpha_3,\alpha_4$.
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