Current in the tube flows from cathode to anode (plate), with the grid in between to act as a control valve for that flow. By making the grid negative with respect to the cathode, we are said to 'bias' the tube, which means that we are establishing how much current flows from cathode to anode with no signal present at the grid and a specific positive Voltage on the anode.
In your example schematic, what appears to be a negative 18Vdc (negative with reference to ground) is applied to the grid to establish this control or bias point. this would need to be obtained from a separate power supply or battery.
In your amp with the resistor between ground and cathode, the bias is obtained in a different way. The current flowing from ground through the resistor to the cathode on its way to the anode creates a Voltage drop across the resistor, making the cathode positive with respect to the grid. This amounts to the same thing as making the grid negative with respect to the cathode. Thus, the tube is biased. The amount of positive Voltage at the cathode will depend on how much current is flowing through that resistor, as Ohm's Law tells us. The advantage to this type of biasing scheme is that it does not require a separate power supply to establish the bias point. The trade-off is that that resistor in the cathode circuit is consuming a bit of power which is no longer available for the output of the amp.
You are correct in your assumption that the cathode current in your example schematic will also flow through the secondary of the OPT and the voice coil of the speaker which is in parallel with it. But, the DC Voltage this produces on that parallel combination is again dictated by Ohm's Law: E= IxR. So, the Voltage (E) is the product of the current (I) and the resistance (R). Now, the parallel combination of the secondary of the OPT and the speaker voice coil is bound to add up to a very small resistance, certainly in the single digits. Given that the current will not even reach single digits (even 1 amp through a 6V6 would be a disaster!), you will see that the DC Voltage across the secondary/voice coil combination will be more in the range of a fraction of a Volt. For instance, in the example you gave of your amp, the 250 Ohm resistance in the cathode circuit would require .048 Amps, or 48 miliAmps to produce a Voltage drop of 12 Volts across the 250 Ohms. Well, the parallel combination of the secondary and the voice coil in the schematic maybe results in, let's guess it at 4 Ohms. So, that same .048 amps would produce a Voltage drop of only 0.192 Volts.
Hope this helps!
