After a bit more research, the definition of $\kappa \rightarrow [\kappa]^{<\omega}_{\alpha,\beta}$ is not the one I stated. This is where the confusion over Rowbottom cardinals stemmed from.

The real definition is for every $f:[\kappa]^{<\omega}\rightarrow \alpha$ there is an $A\in [\kappa]^\kappa$ such that $|f"[A]^{<\omega}|\leq \beta$.

With this new definition, $f:[\kappa]^{<\omega}\rightarrow \omega$, $f(\eta)=\mbox{ length of } \eta$,

contradicts $\kappa \rightarrow [\kappa]^{<\omega}_{\omega,n}$, (as for any $A\in [\kappa]^\kappa$, $f"[A]^{<\omega}=\omega$). So that this never holds.

However, the question still stands, with the old definition in the last post, (which is a bit weaker) does this imply $\kappa$ is Ramsey? (this seems to be a slight generalization of $\kappa\rightarrow (\kappa)^{<\omega}_\omega$)

This is not a complete answer, but clears up some of the confusion.

After a bit more research, the definition of $\kappa \rightarrow [\kappa]^{<\omega}_{\alpha,\beta}$ is not the one I stated. This is where the confusion over Rowbottom cardinals stemmed from.

The real definition is for every $f:[\kappa]^{<\omega}\rightarrow \alpha$ there is an $A\in [\kappa]^\kappa$ such that $|f"[A]^{<\omega}|\leq \beta$.

With this new definition, $f:[\kappa]^{<\omega}\rightarrow \omega$, $f(\eta)=\mbox{ length of } \eta$,

contradicts $\kappa \rightarrow [\kappa]^{<\omega}_{\omega,n}$, (as for any $A\in [\kappa]^\kappa$, $f"[A]^{<\omega}=\omega$). So that this never holds.

However, the question still stands, with the old definition in the last post, (which is a bit weaker) does this imply $\kappa$ is Ramsey? (this seems to be a slight generalization of $\kappa\rightarrow (\kappa)^{<\omega}_\omega$)