How to use this type of photodiode?
I would advice against buying from RadioShack. They may be handy if there's a shop around the corner, but they don't know what they're selling, and you get no information. Here there's some data though, on the product summary page, not on the technical specs (unless you find "Supported languages: English" for a photodiode useful).  

The product summary page says "Phototransistor detector - 20V 25mA", so that will indeed be a phototransistor and not a photodiode, since they won't give you that much current. But you get no datasheet which would give you sensitivity, so you'll have to experiment.
Also these are IR products, so not directly suitable to measure visible light levels (though they may respond a little bit to that).
In the answer to the question you refer to I showed this picture:  



+Vs is your 5 V. When (IR) light hits the phototransistor it will cause a current which in turn will cause a voltage drop across the resistor due to Ohm's Law: V = I \$\times\$ R. So a 1 &micro;A current will cause a 0.5 V drop, and a 10 &micro;A current a 5 V drop, which you'll see as a high level. It won't go higher, because that's your supply voltage.
But 10 &micro;A is not really much for a phototransistor, so you may want to decrease sensitivity by decreasing the resistor value to for instance 10 k&Omega;. That will give you 3.5 V (minimum for a high level) at 350 &micro;A. You'll have to experiment with the value to get the right sensitivity.

You mention placing the resistor on the high side, and you can do it that way, but then the output voltage range will be reversed: you'll get a low voltage with high light levels.