What is meant in this particular context is almost certainly that for large $t$, we can approximate $\begin{pmatrix} a \operatorname{sech} t \\ b\tanh t \\ c \operatorname{sech} t \end{pmatrix} \sim \begin{pmatrix}0 \\ b \\ 0\end{pmatrix} + e^{-t/k} \begin{pmatrix}p \\ q \\ r\end{pmatrix} + o(e^{-t/k})$ for appropriate constants $k$, $p$, $q$, $r$. The characteristic time of the approach to $(0,b,0)$ is then $k$.
(And your $b$ had better be $1$ for the question to make sense).
Hmm ... alternatively, "align with $(0,1,0)$" could mean the process of the direction of the vector approaching the direction of the positive $y$ axis. In that case the relevant approximation would be something like $ \frac{ \sqrt{a^2+c^2} \operatorname{sech} t }{ b \tanh t } \sim e^{-t/k} s + o(e^{-t/k}) $ for constants $k$ and $s$. Again $k$ is the characteristic time.
Here the left-hand side of this represents the angle between the vector and $(0,1,0)$ as seen from the origin, rather than the distance between your vector and $(0,1,0)$. Strictly speaking the left-hand side should arguably be $\tan^{-1}\left(\frac{\sqrt{a^2+c^2}\operatorname{sech}t}{b\tanh t}\right)$, but since $\tan^{-1}(x)\sim x$ for small $x$ anyway, it doesn't matter for the result.
(So much for "almost certainly").