My understanding of the your design is that the entire device is on a single PCB, is within a single enclosure, and is connected to the host by a single USB cable. You've integrated a hub onto the PCB to allow both the devices to communicate with the PC. The following answer will hinge on these assumptions, if it's made of several separate devices connected by disconnectable cables then that changes things.
In this case, I suggest that you simply configure the hub to enumerate as a high-power device, and share the resulting 500 mA among the whole board. Interestingly enough, TI's ganged-port sample schematic shows the devices all connected together, even when using their power management IC:
The incoming 5V power supply line (highlighted in blue, as it's one of two nets that we're interested in on this complicated schematic) is connected to a TPS2041 power management IC (a generous description, it's really just a FET that shuts down when it detects 500mA of current being passed). However, each of the inputs are shorted together, and each of the outputs are shorted together as well, and then distributed to each of the downstream ports (the net shown in red).
Basically, they're doing overcurrent protection for all of the downstream sections in a single IC. They have no way of detecting whether they have three low-power (100mA) units, a single high-power unit, or two low-power units and one 300 mA unit. All these options are acceptable based on this reference design. You wrote:
According to the USB specification, a bus-powered hub can provide only one unit per downstream port while drawing max 5 units...
but, to directly answer your question, this design from Texas Instruments (a USB group member and major implementor) shows that you only have to guarantee that the total current is less than 5 units.
To solve your problem, the rules state (taken from the excellent USB in a nutshell document):
High power bus powered functions will draw all its power from the bus and cannot draw more than one unit load until it has been configured, after which it can then drain 5 unit loads (500 mA Max) provided it asked for this in its descriptor.
If you can guarantee that your driver stage will not begin drawing current until the device has been configured (which might be as simple as a timed delay in the host controller), you can simply wire everything together. Because your entire circuit is on a single PCB and has no user-accessible downstream ports, you can probably also leave out the TPS2041 and simply design the system to not require more than 500 mA of current in any state.
Another benefit of enumerating as a high-power device is improved input voltage specifications. When you have enumerated as a low-power device, the host is only required to produce 4.40 V at the upstream port (which will be lower at your device due to the resistance of the cable). When you have enumerated as a high-power device, the specification guarantees that you'll get 4.75 V, which is more likely to be within the operating range of any 5V components you may be using.
There is a distinct possibility that the overload on the USB port 5V from the computer is related to the fact that you only hooked up one set of the VSS and VDD pins of the PIC16F877A. It is never safe to operate a chip in this manner as there is never any assurance that the internal MCU design even connects the various VDD pins together and the various VSS pins together with an internal low impedance connection. Lack of these external connections can cause some parts internal to the chip to not be powered correctly and result in improper forward biasing of chip junction diodes. Massive amounts of current can flow in these cases.
Also be advised if you rebuild the circuit with new parts that you also need to be adding bypass capacitors between the VSS and VDD of each chip connection.
With regard to the USB ports of your computer...It is likely that you have damaged things so bad that they may not see life again. Take this incident to re-think the process of powering experimental circuitry from your computer's USB port. Check out a place like eBay where you can find dozens of choices for very low cost USB power blocks that can supply 5V from a AC power source (or also from an automobile aux power jack) to power your experimental circuits. In the unfortunate case of a short or some oversight in the design of your circuit it is far far better to blow out a 5->10$ power brick than it is to destroy your laptop or desktop computer.
Best Answer
Nope. The problem is impedance. USB 2.0 is high frequency enough to have signal refections at any point with significant impedance mismatch - and your breakout will have a rather large impedance mismatch.
Remember that impedance largely depends on physical properties like size and distance (to each other and ground), so a "breakout" is never a good idea for any HF transmission lines.
The 6 MHz with USB Full devices may work in some cases with short cables, and some AVR programmers only use USB low speed anyway - those are less sensitive to this problem.