Allow me; it's a four-step process. I'll assume these component values:
B+ = 420V
Bias resistor Rk = 2.2k
Transformer DC primary winding resistance = 2k
Transformer reflected primary impedance = 22k
Step 1: You're using two triodes in parallel, so double the vertical current axis;
Step 2: Draw a cathode load line corresponding to the bias resistor, 2.2k;
Step 3: Draw a DC load line corresponding to the total DC resistance in series with the tube, which is the bias resistor plus the transformer DC resistance, 2k + 2.2k. Where the two lines cross is the bias point.
Step 4: Draw an AC load line through the bias point, corresponding to the total impedance in the anode circuit, which is the reflected impedance plus the DC winding resistance (usually we ignore the DC resistance, but since you want all the detail) 2k + 22k.
Step 5: Profit.
Note: Remember that the horizontal axis only shows you the voltage across the valve itself. In this case the bias point shows about 405V. But there is also about 8.5V across the bias resistor, so the voltage you would measure anode-to-ground is 413V. The original Fender schematic shows 410V so I'm pretty close...
And then you need to remember the load is an inductor, highly reactive, so in reality the load line will trace out an ellipse aligned with the idealized straight line. In this case the bias point is so far to the right that the bottom of that ellips will likely smash into the horizontal axis, which explains why the Fender driver hits cut-off so easily; it has crappy current headroom.