What forward airspeed produces maximum rate-of-climb in helicopters?
There are a couple of things at play here:
First, the increase in overall lift of a helicopter as it correlates to an increase in forward airspeed actually has a lot to do with the establishment of Effective Translational Lift (ETL) or the airspeed at which the entire rotor system realizes the benefit of the horizontal air flow. This happens when the helicopter's rotor disc moves completely out of its own downwash and into undisturbed air.
Second, beyond the ETL improvement, forward airspeed in a helicopter only produces an increase in asymmetrical lift, rather than the the nice, tidy symmetrical lift dispersal of fixed wing aircraft. This is because of a principle known as (surprise, surprise) Dissymmetry of Lift. The forward airspeed is added to the rotational speed of the advancing rotor blade on one side of the aircraft and subtracted from the retreating rotor speed on the opposite side of the aircraft. While this is counteracted to some extent by mechanical adjustments (flapping, feathering, leading-lagging) to equalize the inherent rolling tendency, this means that the forward speed alone, as a catalyst for lift production, is not the be-all and end-all.
You are correct that a helicopter, under the right conditions and possessing a surplus of available power can generate lift at zero airspeed. This is why a helicopter, under these conditions, effectively has a Vx of zero, since its best angle-of-climb over distance would be a straight vertical climb.
For Vy, the best rate-of-climb over time performance, it is important to plan for torque settings at level flight. Climb performance charts show the change in torque, above or below torque, required for level flight under the same gross weight and atmospheric conditions to obtain a given rate of climb or descent.
Here is a sample calculation taken from the FAA Helicopter Flying Handbook:

Determine  torque  setting  for  cruise  or  level  flight  using  Figure 7-7.
Pressure Altitude = 8,000 feet,  Outside Air Temperature = +15 °C,  Indicated Airspeed = 80 knots,  Maximum Gross Weight = 5,000 lb.

With this chart, first confirm that it is for a pressure altitude of 8,000 feet with an OAT of 15°. Begin on the left side at 80 knots indicated airspeed (point A) and move right to maximum gross weight of 5,000 lb (point B). From that point, proceed down to the torque reading for level flight, which is 74 percent torque (point C). This torque setting is used in the next problem to add or subtract cruise/descent torque percentage from cruise flight.
Determine climb/descent torque percentage using Figure 7-8.
Rate of Climb or Descent = 500fpm,  Maximum Gross Weight = 5,000lb.

With this chart, first locate a 500-fpm rate of climb or descent (point  A), and then move to the right to a maximum gross weight of 5,000 lb (point B). From that point, proceed down to the torque percentage, which is 15 percent torque (point C). For climb or descent, 15 percent torque should be added/subtracted from the 74 percent torque needed for level flight. For example, if the numbers were to be used for a climb torque, the pilot would adjust torque settings to 89 percent for optimal climb performance.

As you can see on chart 7.7, the green line represents the Vy speed and is relatively low, because it represents the speed at which the lowest percentage of maximum torque is required to sustain level flight in the given conditions, and thus, the speed at which there is the maximum amount of surplus torque available to climb, all while remaining in ETL flight.