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Here is what I think is a partial answer to the problem.

Assuming that $(P_1,\leq_1),…,(P_n,\leq_n)$ are all subspaces of $(P,\leq)$ and $P_1,\ldots,P_n$ form a weak partition of $P$. If all $n$ spaces have order dimension at most $k$, this does not yield a bound on the order dimension of $(P,\leq)$ in general.

Of course, it does yield an upper bound if we have $k=1$. Then, $P_1,\ldots,P_n$ are chains and this is known to cause that the order dimension of $(P,\leq)$ is at most $k$ (in fact, this is one of the most fundemental facts about the order dimension).

However, as soon as we have $k=2$, this is not true anymore, even if we require that $(P,\leq)$ is connected and has top and bottom. Take, for instance, $P(n) = \{0,a_1,\ldots,a_n,b_1,\ldots,b_n,1\}$ and define $\leq$ to be the order given by $x \leq y$ for $x \neq y$ if and only $y = 1$ or $x=0$ or $x = a_i$ and $y = b_j$ for some $i \neq j$. Then, for each $n \in \mathbb{N}$, $P(n)$ can be covered by three subspaces $(P_1,\leq_1),(P_2,\leq_2),(P_3,\leq_3)$, each of which is a tree. Trees have order dimension $2$, so $P(n)$ can always be covered by three subspaces of order dimension $2$. But now, $(P(n),\leq)$ has order dimension $n$ (in fact, it is the poset that is obtained by adding a greatest and least element to what is often called the standard example of a poset of order dimension $n$).

This, however, does not answer the version with the variant in which the transitive closure of $\bigcup \leq_i$ has to be $\leq$.

Here is what I think is a partial answer to the problem.

Assuming that $(P_1,\leq_1),…,(P_n,\leq_n)$ are all subspaces of $(P,\leq)$ and $P_1,\ldots,P_n$ form a weak partition of $P$. If all $n$ spaces have order dimension at most $k$, this does not yield a bound on the order dimension of $(P,\leq)$ in general.

Of course, it does yield an upper bound if we have $k=1$. Then, $P_1,\ldots,P_n$ are chains and this is known to cause that the order dimension of $(P,\leq)$ is at most $k$ (in fact, this is one of the most fundamental facts about the order dimension).

However, as soon as we have $k=2$, this is not true anymore, even if we require that $(P,\leq)$ is connected and has top and bottom. Take, for instance, $P(n) = \{0,a_1,\ldots,a_n,b_1,\ldots,b_n,1\}$ and define $\leq$ to be the order given by $x \leq y$ for $x \neq y$ if and only $y = 1$ or $x=0$ or $x = a_i$ and $y = b_j$ for some $i \neq j$. Then, for each $n \in \mathbb{N}$, $P(n)$ can be covered by three subspaces $(P_1,\leq_1),(P_2,\leq_2),(P_3,\leq_3)$, each of which is a tree. Trees have order dimension $2$, so $P(n)$ can always be covered by three subspaces of order dimension $2$. But now, $(P(n),\leq)$ has order dimension $n$ (in fact, it is the poset that is obtained by adding a greatest and least element to what is often called the standard example of a poset of order dimension $n$).

This, however, does not answer the version with the variant in which the transitive closure of $\bigcup \leq_i$ has to be $\leq$.

Here is what I think is a partial answer to the problem.

Assuming that $(P_1,\leq_1),…,(P_n,\leq_n)$ are all subspaces of $(P,\leq)$ and $P_1,\ldots,P_n$ form a weak partition of $P$. If all $n$ spaces have order dimension at most $k$, this does not yield a bound on the order dimension of $(P,\leq)$ in general.

Of course, it does yield an upper bound if we have $k=1$. Then, $P_1,\ldots,P_n$ are chains and this is known to cause that the order dimension of $(P,\leq)$ is at most $k$ (in fact, this is one of the most fundemental facts about the order dimension).

However, as soon as we have $k=2$, this is not true anymore, even if we require that $(P,\leq)$ is connected and has top and bottom. Take, for instance, $P(n) = \{0,a_1,\ldots,a_n,b_1,\ldots,b_n,1\}$ and define $\leq$ to be the order given by  $x \leq y$ if and only if $y =1$ or $x=0$ or $x = a_i$ and $y = b_j$ for some $i \neq j$. Then, for each $n \in \mathbb{N}$, $P(n)$ can be covered by three subspaces $(P_1,\leq_1),(P_2,\leq_2),(P_3,\leq_3)$, each of which is a tree. Trees have order dimension $2$, so $P(n)$ can always be covered by three subspaces of order dimension $2$. But now, $(P(n),\leq)$ has order dimension $n$ (in fact, it is the poset that is obtained from adding a greatest element to what is often called the standard example of a poset of order dimension $n$).

This, however, does not answer the version with the variant in which the transitive closure of $\bigcup \leq_i$ has to be $\leq$.

Here is what I think is a partial answer to the problem.

Assuming that $(P_1,\leq_1),…,(P_n,\leq_n)$ are all subspaces of $(P,\leq)$ and $P_1,\ldots,P_n$ form a weak partition of $P$. If all $n$ spaces have order dimension at most $k$, this does not yield a bound on the order dimension of $(P,\leq)$ in general.

Of course, it does yield an upper bound if we have $k=1$. Then, $P_1,\ldots,P_n$ are chains and this is known to cause that the order dimension of $(P,\leq)$ is at most $k$ (in fact, this is one of the most fundemental facts about the order dimension).

However, as soon as we have $k=2$, this is not true anymore, even if we require that $(P,\leq)$ is connected and has top and bottom. Take, for instance, $P(n) = \{0,a_1,\ldots,a_n,b_1,\ldots,b_n,1\}$ and define $\leq$ to be the order given by  $x \leq y$ for $x \neq y$ if and only $y = 1$ or $x=0$ or $x = a_i$ and $y = b_j$ for some $i \neq j$. Then, for each $n \in \mathbb{N}$, $P(n)$ can be covered by three subspaces $(P_1,\leq_1),(P_2,\leq_2),(P_3,\leq_3)$, each of which is a tree. Trees have order dimension $2$, so $P(n)$ can always be covered by three subspaces of order dimension $2$. But now, $(P(n),\leq)$ has order dimension $n$ (in fact, it is the poset that is obtained by adding a greatest and least element to what is often called the standard example of a poset of order dimension $n$).

This, however, does not answer the version with the variant in which the transitive closure of $\bigcup \leq_i$ has to be $\leq$.

Here is what I think is a partial answer to the problem.

Assuming that $(P_1,\leq_1),…,(P_n,\leq_n)$ are all subspaces of $(P,\leq)$ and $P_1,\ldots,P_n$ form a weak partition of $P$. If all $n$ spaces have order dimension at most $k$, this does not yield a bound on the order dimension of $(P,\leq)$ in general.

Of course, it does yield an upper bound if we have $k=1$. Then, $P_1,\ldots,P_n$ are chains and this is known to cause that the order dimension of $(P,\leq)$ is at most $k$ (in fact, this is one of the most fundemental facts about the order dimension).

However, as soon as we have $k=2$, this is not true anymore. Take, for instance, $P(n) = \{ a_1,\ldots,a_n,b_1,\ldots,b_n \}$ and define $\leq$ to be the order given by $x \leq y$ if and only if either $y =1$ or $x = a_i$ and $y = b_j$ for some $i \neq j$. Then, for each $n \in \mathbb{N}$, $P(n)$ can be covered by three subspaces $(P_1,\leq_1),(P_2,\leq_2),(P_3,\leq_3)$, each of which is a tree. Trees have order dimension $2$, so $P(n)$ can always be covered by three subspaces of order dimension $2$. But now, $(P(n),\leq)$ has order dimension $n$ (in fact, it is the poset that is obtained from adding a greatest element to what is often called the standard example of a poset of order dimension $n$).

This, however, does not answer the version with the variant in which the transitive closure of $\bigcup \leq_i$ has to be $\leq$.

Here is what I think is a partial answer to the problem.

Assuming that $(P_1,\leq_1),…,(P_n,\leq_n)$ are all subspaces of $(P,\leq)$ and $P_1,\ldots,P_n$ form a weak partition of $P$. If all $n$ spaces have order dimension at most $k$, this does not yield a bound on the order dimension of $(P,\leq)$ in general.

Of course, it does yield an upper bound if we have $k=1$. Then, $P_1,\ldots,P_n$ are chains and this is known to cause that the order dimension of $(P,\leq)$ is at most $k$ (in fact, this is one of the most fundemental facts about the order dimension).

However, as soon as we have $k=2$, this is not true anymore, even if we require that $(P,\leq)$ is connected and has top and bottom. Take, for instance, $P(n) = \{0,a_1,\ldots,a_n,b_1,\ldots,b_n,1\}$ and define $\leq$ to be the order given by $x \leq y$ if and only if $y =1$ or $x=0$ or $x = a_i$ and $y = b_j$ for some $i \neq j$. Then, for each $n \in \mathbb{N}$, $P(n)$ can be covered by three subspaces $(P_1,\leq_1),(P_2,\leq_2),(P_3,\leq_3)$, each of which is a tree. Trees have order dimension $2$, so $P(n)$ can always be covered by three subspaces of order dimension $2$. But now, $(P(n),\leq)$ has order dimension $n$ (in fact, it is the poset that is obtained from adding a greatest element to what is often called the standard example of a poset of order dimension $n$).

This, however, does not answer the version with the variant in which the transitive closure of $\bigcup \leq_i$ has to be $\leq$.