Why do transmission line equations use current through inductor rather than capacitor?
Remember that although we can have a voltage drop across an element, the current through a device is a single value.  On the other hand, at a node (wire) the voltage is the same for each path, but the currents can differ as long as they sum to 0.

In the diagram you reference:


there are two labeled currents.  \$I(x)\$ is the current entering the differential element of transmission line and going through the inductor and resistor, and \$I(x+\delta x)\$ is the current leaving the differential element of transmission line.

Since charge is conserved, the amount of current entering a node must be equal to the amount leaving it.  The current through the resistor and inductor cannot be different from one side to another since there is nowhere else for the current to go.

However, at the node above the capacitor the current splits.  We know that \$I(x)\$ enters from the left and \$I(x+\delta x)\$ leaves to the right.  That difference in current must come from the capacitor, therefore the current entering the capacitor is simply the difference, \$I(x)-I(x+\delta x)\$.

Applying the capacitor differential equation, \$i(t)=C\frac{\mathrm dV(t)}{\mathrm dt}\$, we get equation 2:

$$
I(x;t) - I(x+\delta x;t)
=(C\delta x)\frac{\mathrm dV(x;t)}{\mathrm dt}
$$

In other words, your intuition was correct: we are actually using the current through the capacitor, not the "current drop across the inductor."