I'm a little stuck here (and should mention that I lack experience with unbounded sets in the transfinite):
Say we have two uncountable regular cardinals $\kappa$ and $\lambda$ where $\kappa < \lambda$. Consider the ordinal product $\kappa \cdot \lambda$ (i.e., $type(\lambda \times \kappa, <)$ with $<$ the lexicographic ordering).
I would like to find out whether or not (or under what circumstances) it's possible to find a sequence $S$ of elements of $\kappa \cdot \lambda$ so that $S$ has cardinality $\kappa$ and yet $S$ is unbounded in $\kappa \cdot \lambda$. It should possible to find an unbounded sequence $\subseteq \kappa \cdot \lambda$ if we allow it to have cardinality $\lambda$, but how to proceed if we do not have as many sequence elements to work with?
My first impression was that it may not be possible to find such $S$, because I failed to see how to provide a bigger element often enough for any given "tuple" $\alpha \in \kappa \cdot \lambda$. For example, if we take $\kappa = \aleph_1$ and $\lambda = \aleph_2$, and I try to climb up high enough towards $\aleph_2 = |\kappa \cdot \lambda|$, in order to close the gap.
Still I'm a little doubtful that this first impression is true. Maybe there is a misunderstanding concerning the ordinal product in the first place?
Remember the definition of $\alpha\cdot\beta$ when $\beta$ is a limit ordinal (e.g., when it is a regular cardinal) is $\sup\{\alpha\cdot\gamma\mid\gamma<\beta\}$.
Therefore, $\kappa\cdot\lambda=\sup\{\kappa\cdot\alpha\mid\alpha<\lambda\}$. If $\alpha<\lambda$, then $|\kappa\cdot\alpha|<\lambda$ and therefore, by the fact that $\lambda$ is a cardinal, $\alpha\leq\kappa\cdot\alpha<\lambda$. Therefore, $\sup\{\kappa\cdot\alpha\mid\alpha<\lambda\}=\lambda$, so $\kappa\cdot\lambda=\lambda$, and since $\lambda$ is regular, there is no unbounded set of size $\kappa$.
This will also give you a natural cofinal sequence of order type $\lambda$ (even if $\kappa\cdot\lambda\neq\lambda$, that is).