How Were secp*k1 elliptic curve generators chosen?
Well, I'm not privy to how Certicom selected these parameters, however I can tell you that it doesn't really matter; if the generator they selected is weak, then all generators on that curve are weak.

In particular (and assuming that the curve has a prime number of points on it; this is true for the Certicom curves):


If you can solve the discrete log to a specific generator $G$, you can solve it to any base (by solving two instances of the generator $G$ problem)
If you can solve the computational Diffie Hellman problem with a specific generator $G$, you can solve it to any base (by solving $\log p$ instances of the generator $G$ problem).


Hence, if there is a weakness with their generator, then their curve is weak (except possibly against the decisional Diffie-Hellman problem; I don't know of a similar reduction there).

Demonstrating the first reduction is quite straight-forward; suppose thatwe are  given an Oracle, that, given $aG$, recovers $a$ (with $G$ being our distinguished generator).  Then, given an instance which consists of $H$, and $aH$, with $H$ being an arbitrary point, we call the Oracle twice, once with $H=bG$, and once with $aH = cG$, recovering $b$ and $c$.  Then we can compute $a = b^{-1}c \bmod q$ (where $q$ is the order of the curve); this is our answer.

The computational Diffie-Hellman case is somewhat more complex.  For that, we suppose that we have an Oracle that, given $aG$ and $bG$, computes the value $abG$.  Then, the problem we are trying to solve is, given $H$, $aH$ and $bH$, compute the value $abH$.

To see how to we proceed, let us designate $c$ as the value $H = cG$.  Now, we don't know the value of $c$, however we can compute the point $c^2G$ (by using our Diffie-Hellman Oracle, with the inputs $cG$ and $cG$).  Extending this observation, we can compute $c^kG$ with $\log{k}$ calls to the Oracle.  So, we compute $c^{q-2}G = c^{-1}G$ (with $\log{q}$ Oracle calls, and $q$ is the curve order).  Then, with $aH = acG$ and $bH = bcG$, we can compute $abH = abcG$ with two more Oracle calls.