TL;DR: It depends on the power source. Assuming the power is well regulated (i.e. 5V with 5% tolerance either way, not too much fluctuation/rippling1), it's perfectly safe.
For the full explanation, first, we need to delve into how a USB connector/cable is wired.
With a basic USB 2.0 cable, you have a ground (GND) and a +5V (VBUS) for power. You also have a D- and D+ for data. I'm not going to go into the data lines here, since the question is addressing power.
There's really two and a half types of USB Y cables. Physically, there's:
One male connector, two female connectors
There are two types of the one male two female cables (hence the two and a half). One is a basic charging cable, splitting one port's power across two devices. The other is a specialised cable that somehow splits the data lines - and it's only used in very specific applications.
Two male connectors, one (fe)male connector
This is the one you are referring to. What happens is you have your USB cable, one side for the host and one for the device, but with an extra connector attached. This extra connector does not have any data lines (D-, D+); it only has the power lines (GND, VBUS). It's attached in parallel to the existing cable. In other words, VBUS is connected to VBUS and GND is connected to GND.
VBUS -------/ |
D- ---- |
D+ ---- |
What happens is the voltage becomes the average of the two and the max current is the sum of what the host connector and power connector can provide. Assuming both are connected to standard USB power supplies, the combined VBUS line should be at +5V.
Now, what does this mean? Well, max current is the max available, you can draw up to this amount. In other words, the max current provided by the sum of the connectors should be at least what your device needs. As a rule of thumb, being able to provide more power is not a bad thing. The voltage, for USB, must be at 5V.
So, if you have a computer connected to the host connector, the USB 2.0 specifications limit a standard port to 500mA. This may not be enough for higher power devices, in which case you can connect the power connector to an additional source, often another port on the computer. Depending on the port2, you may get up to 500mA again. If you add them, you get 1000mA at 5V, providing more power for the device to draw on.
Now, your question asks about an external power source. Assuming a good USB charger, and no shorts anywhere, no damage will result: the voltage will be 5 V and the max current will add to whatever - but the device will only draw as much as it needs.
It gets tricky, however. There are cheap USB chargers out there with horrible regulation and a wide variance in voltage, especially depending when loaded. This may not be a major issue with some devices, even possibly including phones which may have their own internal regulation. However, other devices may be more sensitive. It depends on how bad the power source is, and how sensitive the device is; both are really a case-by-case thing, though some device categories will perform better than others.
Generally, if you have a reliable USB power source/charger, everything will be perfectly fine. A not-so-reliable one might work, but, well, won't be very reliable.
1The USB specifications specify 5% tolerance for USB 2.0, i.e. from 4.75V to 5.25V. The actual device you are using may be more lenient.
2A standard port should only provide up to 100mA without negotiation. Since a power connector has no data lines, it cannot request more power. So typically the host + power connectors can provide a max of 600mA. However, there are special charging ports in the specification that may provide 500mA or even more without negotiation, and those ports may actually be more common nowadays. This isn't really relevant to using an external power source, however.