The purpose of the coupling capacitors is to block DC. A 10nF capacitor looks like an open circuit to DC voltages and a short circuit to high frequency (GHz) AC voltages. The reason that DC needs to be blocked is that the DC bias levels on either side of the SATA interface would be different. The DC bias is the quiescent voltage level of the Tx/Rx pins in the absence of a signal. Without the capacitors, the DC bias of the MCU would disturb the DC bias of the bridge IC. BTW, after removing the capacitors, there will still be coupling capacitors on the computer's motherboard to provide DC isolation.
Perhaps the following example will better illustrate what is happening.
Code:
+12V +50V
| |
| |
|\| 6V 0V |\|
___| \_____||_____| \___
| / || | /
|/| C |/|
| |
| |
0V -50V
preamplifier power amplifier
The preamplifier has a 12V supply. Its output is biased at the midpoint of the supply rail, namely 6V. The AC signal then swings symmetrically above and below this bias point. For example, a 1Vpp (peak-to-peak) AC signal would swing between 5.5V and 6.5V.
The power amp has a +/-50V supply. Its input has a DC bias of 0V. The coupling capacitor, C, blocks the DC component of the input signal, resulting in a swing of -0.5V to +0.5V.