Contradiction between complex baseband and real-valued baseband
As a sidenote:

When you demodulate your signal $x_{\mathrm{rf}}(t)$  you say that there is some phase difference $\phi$. This translates to crosstalk (i.e., the quadrature and in-phase components influence each other at the receiver). It's easy to see that if $\phi = 0$ (namely, we know exactly when to sample the received signal), there is no crosstalk at all:

$$y_i = \sum_{k=0}^K \left( g_k x_i[n-k] + g_k \tan(0) x_q [n-k] \right) =  \sum_{k=0}^K  g_k x_i[n-k]$$

In the baseband domain, this means that the filter $c_k$ is purely real (if it was purely imaginary, then there wouldn't be crosstalk either, you just have to invert the channels to retrieve the information correctly).

The answer:

Whether there is crosstalk or not, both of your equations get to the same result, but with different names. Using your notation, $a_k$ in the first equation would be equivalent to $g_k$ in the second one, and $b_k$ to $-g_k\tan(\phi)$. I believe that your confusion arose from thinking of $a_k$ and $b_k$ as two different filters (which is correct), instead of thinking of them as the real and imaginary part of a single filter ($c_k$).