With another clockwise fan from the same maker, I was able to reverse the direction by interchanging the yellow and black wires as some answers here indicated. The explanation as I understood is that the rewiring changes the winding with which the capacitor is in series and hence the starting direction is inverted.

In 3-phase motors, each of the three stator windings carry a current out of phase with others and the phase difference generates the rotating magnetic field required to cause the motion. With single-phase ones, a phase difference is engineered by splitting the single phase current into two stator windings and putting a capacitor in series with one of the windings so that there's a 90° phase difference between the currents in the two windings. This page on electric motors explains the concept with illuminating animations.

The following figure shows my guess, based on the above information, at the internal wiring of the clockwise spinning fan whose image is posted in the question, for clockwise and anti-clockwise rotations.

A point to note here is that single-phase AC itself produces a changing magnetic field - though a pulsating one, not a rotating one. But this pulsating field can be resolved, as per the double field revolving theory (the link has an excellent video of the workings by the way), into two revolving fields rotating in opposite direction to each other. These two fields produce an equal but opposite torque. On a static rotor, they'll cancel each other out. But an initial rotation makes torque in one direction greater than the other and starts up the fan.

This is what, I believe, happened when the OP switched red and yellow wires on his fan. The result was that the capacitor was in series with both windings => there was no phase difference in the currents in the windings. When he added a slight initial rotation, the fan continued spinning in the nudged direction.

Column C

First off you'll notice the text "Column C to be used in all cases except as otherwise permitted in Note 3.)", in the title of table 220.55. This makes it simple. You have 2 appliances, so follow that over in the table, and you'll see 11 in Column C. So there you go, you can just use 11 kW. Done.

11,000 W / 240 V = 45.8333 A

So you'll need a 50 ampere breaker, and wire sized appropriately for the load.

Note 3

Note 3 says:

3. Over 1 3⁄4 kW through 8 3⁄4 kW. In lieu of the method provided in Column C, it shall be permissible to add the nameplate ratings of all household cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW and multiply the sum by the demand factors specified in Column A or Column B for the given number of appliances. Where the rating of cooking appliances falls under both Column A and Column B, the demand factors for each column shall be applied to the appliances for that column, and the results added together.

Perfect, so instead of just using the value from column C you can do math. Let's step through it.

...it shall be permissible to add the nameplate ratings of all household cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW...

8.4 kW + 5.0 kW = 13.4 kW

...and multiply the sum by the demand factors specified in Column A or Column B for the given number of appliances...

Let's check the table again... You have 2 appliances, both between 3 1/2 and 8 3/4 kW. So You'll look at column B, and find 65%.

13.4 kW * 65% = 8.71 kW

8710 W / 240 V = 36.2916 A

So using this method you can use a 40 ampere breaker, and appropriately sized wire. However, keep in mind that if you change the equipment, you'll have to do the calculation again. So while you can use this value, you may have to upgrade the circuit later if you change equipment.

Note 4

I'm not exactly sure how note 4 comes into play, but I think it can be used if this is the only equipment on the branch circuit. Just for fun, let's run through that one too.

4. Branch-Circuit Load. It shall be permissible to calculate the branch-circuit load for one range in accordance with Table 220.55. The branch-circuit load for one wall-mounted oven or one counter-mounted cooking unit shall be the nameplate rating of the appliance. The branch-circuit load for a counter-mounted cooking unit and not more than two wall-mounted ovens, all supplied from a single branch circuit and located in the same room, shall be calculated by adding the nameplate rating of the individual appliances and treating this total as equivalent to one range.

You're only concerned with the second half of this note, since you have one counter-mounted cooking unit, and one wall-mounted oven, all supplied by a single branch-circuit, and located in the same room. So you can add the nameplate values, and treat it as a single range.

8.4 kW + 5.0 kW = 13.4 kW

So you can treat the units as a single 13.4 kW range. Check the column C again, this time for a single range. You'll find a value of 8 kW. But wait... The column header says "(Not over 12 kW Rating)". Your range is 13.4 kW. That's bigger than 12 kW. Now you'll have to check note 1

1. Over 12 kW through 27 kW ranges all of same rating. For ranges individually rated more than 12 kW but not more than 27 kW, the maximum demand in Column C shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kW.

That's easy enough.

13.4 kW - 8 kW = 5.4 kW

Since .4 is not a "major fraction", you can just use 5 kW. So you'll have to add 5% 5 times.

5% * 5 = 0.25
8000 W * 0.25 = 2000 W
8000 W + 2000 W = 10,000 W

That means you'll have to use 10 kW as your demand.

10,000 W / 240 V = 41.666 A

Which means you can use a 50 ampere breaker, and appropriately size wire.