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Theorem 1 is due to P. Erdős and E. Specker, On a theorem in the theory of relations and a solution of a problem of Knaster, Colloq. Math. 8 (1961), 19–21 (pdf).

Your proposed common generalization of Theorems 1 and 2, specialized to $\eta=\omega+1$, contradicts the continuum hypothesis, or more precisely the "stick" principle.

Theorem. Assuming CH, there is a mapping $f:\omega_1\to\mathcal P(\omega_1)$ such that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and such that $\omega_1$ is not the union of countably many $f$-free sets.

Proof. Assuming CH we can construct the mapping $f$ so that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and for each subset $X$ of $\omega_1$ of order type $\omega$ we have $X=f(\alpha)$ for some $\alpha\lt\omega_1$.

Assume for a contradiction that $\omega_1=\bigcup_{n\lt\omega}A_n$ where each $A_n$ is $f$-free. Choose an ordinal $\beta\lt\omega_1$ so that, for each $n\lt\omega$, either $A_n\subseteq\beta$ or else $A_n$ is uncountable. Construct a set $X\subseteq\omega_1$ of order type omega such that $X\cap(A_n\setminus\beta)\ne\varnothing$ whenever $A_n$ is uncountable, and let $X=f(\alpha)$. Then $\alpha\in A_n$ for some $n$, and $A_n$ is uncountable since $\alpha\gt\beta$, so $f(\alpha)\cap A_n\ne\varnothing$, contradicting our assumption that $A_n$ is $f$-free.

Remark. So the CH (in fact the combinatorial principle "stick" which is weaker) implies that the common generalization is false for $\eta=\omega+1$. On the other hand, $\text{MA}_{\aleph_1}$ implies that the common generalization is true for any $\eta\lt\omega_1$.

Theorem 1 is due to P. Erdős and E. Specker, On a theorem in the theory of relations and a solution of a problem of Knaster, Colloq. Math. 8 (1961), 19–21 (pdf). This Erdős–Specker paper was cited by P. Erdős, A. Hajnal, and E. C. Milner, Set mappings and polarized partition relations, Colloq. Math. Soc. János Bolyai 4 (1969), 327–363 (pdf).

Your proposed common generalization of Theorems 1 and 2, specialized to $\eta=\omega+1$, contradicts the continuum hypothesis, or more precisely the "stick" principle.

Theorem. Assuming CH, there is a mapping $f:\omega_1\to\mathcal P(\omega_1)$ such that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and such that $\omega_1$ is not the union of countably many $f$-free sets.

Proof. Assuming CH we can construct the mapping $f$ so that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and for each subset $X$ of $\omega_1$ of order type $\omega$ we have $X=f(\alpha)$ for some $\alpha\lt\omega_1$.

Assume for a contradiction that $\omega_1=\bigcup_{n\lt\omega}A_n$ where each $A_n$ is $f$-free. Choose an ordinal $\beta\lt\omega_1$ so that, for each $n\lt\omega$, either $A_n\subseteq\beta$ or else $A_n$ is uncountable. Construct a set $X\subseteq\omega_1$ of order type omega such that $X\cap(A_n\setminus\beta)\ne\varnothing$ whenever $A_n$ is uncountable, and let $X=f(\alpha)$. Then $\alpha\in A_n$ for some $n$, and $A_n$ is uncountable since $\alpha\gt\beta$, so $f(\alpha)\cap A_n\ne\varnothing$, contradicting our assumption that $A_n$ is $f$-free.

Remark. So the CH (in fact the combinatorial principle "stick" which is weaker) implies that the common generalization is false for $\eta=\omega+1$. On the other hand, $\text{MA}_{\aleph_1}$ implies that the common generalization is true for any $\eta\lt\omega_1$.

Theorem 1 is due to P. Erdős and E. Specker, On a theorem in the theory of relations and a solution of a problem of Knaster, Colloq. Math. 8 (1961), 19-21 (pdf).

Your proposed common generalization of Theorems 1 and 2, specialized to $\eta=\omega+1$, contradicts the continuum hypothesis, or more precisely the "stick" principle.

Theorem. Assuming CH, there is a mapping $f:\omega_1\to\mathcal P(\omega_1)$ such that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and such that $\omega_1$ is not the union of countably many $f$-free sets.

Proof. Assuming CH we can construct the mapping $f$ so that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and for each subset $X$ of $\omega_1$ of order type $\omega$ we have $X=f(\alpha)$ for some $\alpha\lt\omega_1$.

Assume for a contradiction that $\omega_1=\bigcup_{n\lt\omega}A_n$ where each $A_n$ is $f$-free. Choose an ordinal $\beta\lt\omega_1$ so that, for each $n\lt\omega$, either $A_n\subseteq\beta$ or else $A_n$ is uncountable. Construct a set $X\subseteq\omega_1$ of order type omega such that $X\cap(A_n\setminus\beta)\ne\varnothing$ whenever $A_n$ is uncountable, and let $X=f(\alpha)$. Then $\alpha\in A_n$ for some $n$, and $A_n$ is uncountable since $\alpha\gt\beta$, so $f(\alpha)\cap A_n\ne\varnothing$, contradicting our assumption that $A_n$ is $f$-free.

Remark. So the CH (in fact the combinatorial principle "stick" which is weaker) implies that the common generalization is false for $\eta=\omega+1$. On the other hand, $\text{MA}_{\aleph_1}$ implies that the common generalization is true for any $\eta\lt\omega_1$.

Theorem 1 is due to P. Erdős and E. Specker, On a theorem in the theory of relations and a solution of a problem of Knaster, Colloq. Math. 8 (1961), 19–21 (pdf).

Your proposed common generalization of Theorems 1 and 2, specialized to $\eta=\omega+1$, contradicts the continuum hypothesis, or more precisely the "stick" principle.

Theorem. Assuming CH, there is a mapping $f:\omega_1\to\mathcal P(\omega_1)$ such that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and such that $\omega_1$ is not the union of countably many $f$-free sets.

Proof. Assuming CH we can construct the mapping $f$ so that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and for each subset $X$ of $\omega_1$ of order type $\omega$ we have $X=f(\alpha)$ for some $\alpha\lt\omega_1$.

Assume for a contradiction that $\omega_1=\bigcup_{n\lt\omega}A_n$ where each $A_n$ is $f$-free. Choose an ordinal $\beta\lt\omega_1$ so that, for each $n\lt\omega$, either $A_n\subseteq\beta$ or else $A_n$ is uncountable. Construct a set $X\subseteq\omega_1$ of order type omega such that $X\cap(A_n\setminus\beta)\ne\varnothing$ whenever $A_n$ is uncountable, and let $X=f(\alpha)$. Then $\alpha\in A_n$ for some $n$, and $A_n$ is uncountable since $\alpha\gt\beta$, so $f(\alpha)\cap A_n\ne\varnothing$, contradicting our assumption that $A_n$ is $f$-free.

Remark. So the CH (in fact the combinatorial principle "stick" which is weaker) implies that the common generalization is false for $\eta=\omega+1$. On the other hand, $\text{MA}_{\aleph_1}$ implies that the common generalization is true for any $\eta\lt\omega_1$.

Partial answer. Your proposed common generalization of Theorems 1 and 2, specialized to $\eta=\omega+1$, contradicts the continuum hypothesis, or more precisely the "stick" principle.

Theorem. Assuming CH, there is a mapping $f:\omega_1\to\mathcal P(\omega_1)$ such that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and such that $\omega_1$ is not the union of countably many $f$-free sets.

Proof. Assuming CH we can construct the mapping $f$ so that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and for each subset $X$ of $\omega_1$ of order type $\omega$ we have $X=f(\alpha)$ for some $\alpha\lt\omega_1$.

Assume for a contradiction that $\omega_1=\bigcup_{n\lt\omega}A_n$ where each $A_n$ is $f$-free. Choose an ordinal $\beta\lt\omega_1$ so that, for each $n\lt\omega$, either $A_n\subseteq\beta$ or else $A_n$ is uncountable. Construct a set $X\subseteq\omega_1$ of order type omega such that $X\cap(A_n\setminus\beta)\ne\varnothing$ whenever $A_n$ is uncountable, and let $X=f(\alpha)$. Then $\alpha\in A_n$ for some $n$, and $A_n$ is uncountable since $\alpha\gt\beta$, so $f(\alpha)\cap A_n\ne\varnothing$, contradicting our assumption that $A_n$ is $f$-free.

Remark. So the CH (in fact the combinatorial principle "stick" which is weaker) implies that the common generalization is false for $\eta=\omega+1$. On the other hand, $\text{MA}_{\aleph_1}$ implies that the common generalization is true for any $\eta\lt\omega_1$.

Theorem 1 is due to P. Erdős and E. Specker, On a theorem in the theory of relations and a solution of a problem of Knaster, Colloq. Math. 8 (1961), 19-21 (pdf).

Your proposed common generalization of Theorems 1 and 2, specialized to $\eta=\omega+1$, contradicts the continuum hypothesis, or more precisely the "stick" principle.

Theorem. Assuming CH, there is a mapping $f:\omega_1\to\mathcal P(\omega_1)$ such that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and such that $\omega_1$ is not the union of countably many $f$-free sets.

Proof. Assuming CH we can construct the mapping $f$ so that $f(\alpha)\subseteq\alpha$ and $\operatorname{tp}f(\alpha)\le\omega$ for all $\alpha\lt\omega_1$, and for each subset $X$ of $\omega_1$ of order type $\omega$ we have $X=f(\alpha)$ for some $\alpha\lt\omega_1$.

Assume for a contradiction that $\omega_1=\bigcup_{n\lt\omega}A_n$ where each $A_n$ is $f$-free. Choose an ordinal $\beta\lt\omega_1$ so that, for each $n\lt\omega$, either $A_n\subseteq\beta$ or else $A_n$ is uncountable. Construct a set $X\subseteq\omega_1$ of order type omega such that $X\cap(A_n\setminus\beta)\ne\varnothing$ whenever $A_n$ is uncountable, and let $X=f(\alpha)$. Then $\alpha\in A_n$ for some $n$, and $A_n$ is uncountable since $\alpha\gt\beta$, so $f(\alpha)\cap A_n\ne\varnothing$, contradicting our assumption that $A_n$ is $f$-free.

Remark. So the CH (in fact the combinatorial principle "stick" which is weaker) implies that the common generalization is false for $\eta=\omega+1$. On the other hand, $\text{MA}_{\aleph_1}$ implies that the common generalization is true for any $\eta\lt\omega_1$.