NOTE: The original question is about the definition of multichains, as mentioned in the statement. I’ve been given confirmation that the definition indeed means what I thought it is, so now I’m turning towards proving the statement. Please have a look at “Edit 1” and “Edit 2” for my attempt.
I’m reading a text with the following remark:
The order-preserving maps $\phi$ from a poset $\Pi$ to the set $[n]$ is in bijection with the multichains of order ideals of $\Pi$ of length $n$.
A multichain is defined as a sequence of comparable elements, where we allow repetition. I find this definition sort of vague, and couldn’t find another source that use the word “multichain” in the same way.
Could anyone point out, in the example below (where $\phi$ is an order-preserving map), what the second object in the remark is? 
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Edit:
I’m following this book (page 29-30), where order ideals are defined as follows:
Let $\phi: \Pi \rightarrow[2]$ be an order-preserving map into the 2 -chain, and let $I:=\phi^{-1}(1)$. Now $$y \in I \text { and } x \preceq_{\Pi} y \quad \Longrightarrow \quad x \in I. $$ A subset $I \subseteq \Pi$ with this property is called an order ideal of $\Pi$.
According to this definition, I guess the order ideals of the poset above and its Hasse diagram (i.e. I’m drawing $(\mathcal{I}(\Pi), \subseteq)$) are those in this picture?
Furthermore, the statement I’m asking about is Proposition 2.1.2. in the book (page 30), which says that the order ideals (of $(\mathcal{I}(\Pi), \subseteq)$) that we are interested in are of the form: $\emptyset = I_0 \subseteq I_1 \subseteq \dots \subseteq I_n = \Pi$. So I think the multichain we are interested in is the one in this picture?
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Edit 2: I still have to show that the mapping I’ve found is injective (to be honest I’m a bit stuck here). Please have a look at my attempt so far.
Given $(\Pi, \leq_\Pi)$, let $M_\Pi(n)$ be the set of multichains of order ideals $\emptyset = I_0 \subseteq I_1 \subseteq \cdots \subseteq I_n = \Pi$ of length $n$. Let $\varphi: \Omega_\Pi(n) \to M_\Pi(n)$ be defined as $\varphi(\phi) = \{\emptyset, \phi^{-1}(1),\phi^{-1}(2),\dots,\phi^{-1}(n)\}$, where $\Omega_\Pi(n)$ is the set of order-preserving maps $\phi: \Pi \to [n]$.
We claim $\varphi$ is a bijection from $\Omega_\Pi(n)$ to $M_\Pi(n)$.
We check for $\underline{\text{well-definedness}}$ of $\varphi$: given $\phi \in \Omega_\Pi(n)$, let $m_1, m_2 \in M_\Pi(n)$ such that $\phi$ is mapped to both $m_1$ and $m_2$ by $\varphi$. For $i \in [n]$, $\phi^{-1}(i) \in m_1$ and $\phi^{-1}(i) \in m_2$, which shows that $m_1 = m_2$.
$\underline{\text{Surjectivity:}}$ We want to show that, for all $m \in M_\Pi(n)$, $m = \{\emptyset, \phi^{-1}(1),\phi^{-1}(2),\dots,\phi^{-1}(n)\} = \{I_0,I_1,\dots,I_n\}$, where $\emptyset = I_0 \subseteq I_1 \subseteq \dots \subseteq I_n = \Pi$, there exists $\phi \in \Omega_\Pi(n)$ such that $\varphi(\phi) = m$.
We define $\phi: (\Pi,\leq_\Pi) \to ([n],\leq_{\mathbb{N}^*})$ recursively: for $i = n,\dots,1, \ \forall z \in I_i \setminus I_{i-1}, \text{ define } \phi(z) = i$.
This map is well-defined. Let $z \in \Pi$ such that $z$ is mapped to both $i$ and $j \in [n]$ by $\phi$. If $\phi$ is a constant function, then $i = j$, and we are done. Otherwise, assume for a contradiction that $i < j$. Then since $z \mapsto j, \ z \in I_j \setminus I_{j-1}$ by definition. If $I_i \subseteq I_{j-1}$, then $z \notin I_i \Rightarrow z \not\mapsto i, \Rightarrow \Leftarrow$. Otherwise, we have $I_i \supset I_{j-1} \Rightarrow I_i \supseteq I_j$. If $I_i = I_j$ (with $i < j$), then the only possible case is for $I_i = I_j = \emptyset$, which is not true since $z \in I_j$. If $I_i \supset I_j$, then $z \notin I_j \Rightarrow z \not\mapsto j, \Rightarrow \Leftarrow$. Arguing similarly, we have $i \ngtr j$.
$\phi$ is also order-preserving. Let $z_1 \leq_\Pi z_2$, and $\phi(z_1) = i_1, \phi(z_2) = i_2$. We want to show $i_1 \leq_{\mathbb{N}^*} i_2$. Now since $\phi(z_2) = i_2$, $z_2 \in I_{i_2} \setminus I_{{i_2}-1}$. In particular, $z_2 \in I_{i_2}$, and since $I_{i_2}$ is an order ideal and $z_1 \leq_\Pi z_2, \ z_1 \in I_{i_2}$. If $z_1 \in I_{i_2} \setminus I_{i_2 - 1}$, then $i_2 = \phi(z_1) = i_1$. If $z_1 \in I_{i_2 -1}$, then there exists $k \in [i_2 - 1]$ such that $z_1 \in I_k \setminus I_{k-1}$, and so $i_1 = \phi(z_1) = k \leq_{\mathbb{N}^*} i_2$. In both cases, we have $i_1 \leq_{\mathbb{N}^*} i_2$.
Now for all $i \in [n]$, $\phi^{-1}(i) = z \in I_j, \ \forall i \leq_{\mathbb{N}^*} j$ (since $(I_i \setminus I_{i-1}) \subseteq (I_j \setminus I_{i-1})$). Hence we have $\varphi(\phi) = \{I_n,\dots,I_1, \emptyset \} = m$.
Hence we indeed have $\phi \in \Omega_\Pi(n)$.
$\underline{\text{Injectivity:}}$