Let's go through one example, keeping the notations of Marc van Leeuwen's answer you linked to. So let
$G = SU(3) =$ unitary matrices in $GL_3(\Bbb{C})$ with determinant $1$,
$H = \{\pmatrix{a&0&0\\
0&b&0\\
0&0&c\\} \in G\}, $ i.e. $abc=1$ and $a\bar a = b \bar b = c\bar c=1$.
Note that the last condition means $a, b, c\in e^{i\Bbb{R}}$.
We have
$\mathfrak{g} = \mathfrak{su}_3 =$ skew-hermitian matrices in $M_3(\Bbb{C})$ with trace $0$,
$\mathfrak{h} = \{\pmatrix{r&0&0\\
0&s&0\\
0&0&t\\} \in \mathfrak{g}\}, $ i.e. $r+s+t=0$ and $r,s,t \in \Bbb{R}$.
What is the weight lattice $\def\Hom{\operatorname{Hom}}\Lambda = \Lambda_{weight} = \Hom(H, \Bbb C^\times)$ (Hom. of Lie groups/algebraic groups) in this case? It is spanned e.g. by
$$\lambda_1: \pmatrix{a&0&0\\
0&b&0\\
0&0&c\\} \mapsto a$$
$$\lambda_2: \pmatrix{a&0&0\\
0&b&0\\
0&0&c\\} \mapsto b,$$
and actually $\Lambda = \lambda_1^\Bbb{Z}\times\lambda_2^\Bbb{Z}$.
For $\lambda \in \Lambda$, call $\lambda' = \frac{D\lambda}{2\pi i}:\mathfrak{h} \rightarrow \Bbb R$ and notice that we have
$$\lambda'_1: \pmatrix{r&0&0\\
0&s&0\\
0&0&t\\} \mapsto r$$
$$\lambda'_2: \pmatrix{r&0&0\\
0&s&0\\
0&0&t\\} \mapsto s$$
Identifying $\lambda$ with $\lambda'$, we write $\Lambda = \Bbb Z \lambda'_1 \oplus \Bbb Z \lambda'_2$.
The usual simple roots, however, are
$$\alpha_1: \pmatrix{r&0&0\\
0&s&0\\
0&0&t\\} \mapsto r-s$$
$$\alpha_2: \pmatrix{r&0&0\\
0&s&0\\
0&0&t\\} \mapsto s-t,$$
and $\Lambda_{root} = \langle \Phi \rangle = \Bbb Z \alpha_1 \oplus \Bbb Z \alpha_2$.
To see how the two lattices are related, note that $\alpha_1 = \lambda'_1-\lambda'_2$ and, because $r+s+t=0$, we have $\alpha_2 = \lambda'_1+2\lambda'_2$. This gives $\lambda'_1 \equiv \lambda'_2$ mod $\langle \Phi\rangle$ and $3\lambda'_1=2\alpha_1+\alpha_2 \in \langle\Phi\rangle$, but $\lambda'_1 \notin \langle\Phi\rangle$, hence
$$\Lambda_{weight}/\Lambda_{root} = \Lambda/\langle \Phi \rangle \simeq \Bbb Z/3$$
which abstractly tells us that the centre of $G$ will also be isomorphic to $\Bbb Z/3$. However, as is explained in the answer, the above actually identifies $\Lambda/\langle \Phi \rangle$ with the dual of $Z(G)$. To find $Z(G)$ itself, we follow the fourth paragraph of the answer and find that here,
$$\Lambda^{\vee} = \Lambda'^{-1}(\Bbb Z) = \pmatrix{\Bbb Z&0&0\\
0&\Bbb Z&0\\
0&0&\Bbb Z\\} \subset \mathfrak{h}$$
whereas
$$\langle \Phi \rangle^{\vee} = \{h=\pmatrix{r&0&0\\
0&s&0\\
0&0&t\\} \in \mathfrak{h}: \alpha_1(h)\in \Bbb{Z} \ni \alpha_2(h)\}$$
which is computed to be
$$\langle \Phi \rangle^{\vee} = \{\pmatrix{r&0&0\\
0&s&0\\
0&0&t\\} \in \mathfrak{h}: r\in \displaystyle \frac{1}{3}\Bbb Z, r-s \in \Bbb Z \}$$
so that
$$\langle \Phi \rangle^{\vee}/\Lambda^{\vee} = \{\pmatrix{1/3&0&0\\
0&1/3&0\\
0&0&1/3\\}, \pmatrix{2/3&0&0\\
0&2/3&0\\
0&0&2/3\\}, \pmatrix{1&0&0\\
0&1&0\\
0&0&1\\} \}$$
which finally, via the exponential, identifies with the actual centre
$$Z(G) = \{\pmatrix{e^{\frac{2\pi i}{3}}&0&0\\
0&e^{\frac{2\pi i}{3}}&0\\
0&0&e^{\frac{2\pi i}{3}}\\}, \pmatrix{e^{\frac{4\pi i}{3}}&0&0\\
0&e^{\frac{4\pi i}{3}}&0\\
0&0&e^{\frac{4\pi i}{3}}\\}, \pmatrix{1&0&0\\
0&1&0\\
0&0&1\\} \}.$$
Finally, note that all this generalises to $SU(n)$, where one has $\lambda_1, ..., \lambda_{n-1}$ spanning the weight lattice $\Lambda$, $\alpha_1 = \lambda_1-\lambda_2, ..., \alpha_{n-2} = \lambda_{n-2}-\lambda_{n-1}, \alpha_{n-1} = \lambda_1+...+2\lambda_{n-1}$ spanning the root lattice $\langle \Phi \rangle$, and can compute
$$\Lambda/\langle\Phi\rangle \simeq \Bbb Z/n$$
but more precisely
$$\langle \Phi \rangle^{\vee}/\Lambda^{\vee} = \{ k\cdot diag\left(\frac{1}{n},...,\frac{1}{n}\right): k=1, ..., n \}$$
and
$$Z(G) = exp(2\pi i(\langle \Phi \rangle^{\vee}/\Lambda^{\vee})) = \{ diag(\xi^k,...,\xi^k): k=1, ..., n \}$$ with $\xi$ a primitive $n$-th root of unity.
