You need to connect two wires to the (+) side of your power supply, and two wires to the (-) side. If your physical supply has an output cable with only one pair of wires, you'll need to splice the wires together somehow. Connect all the (+) wires together in one group, and the (-) wires in another group.
There are lots of ways to do this. There is a related question with a lot of great answers here on another Stack Exchange site.
A common solution, if you have the equipment, is to solder the wires together and cover the exposed metal with heat-shrink tubing:
An easier option, if your wires are think enough, is to use wire nuts, which can be a good solution for a self-described "beginner" :) You can get them at any hardware store:
A better, but more expensive, option are IDC-style butt-connectors:
That image is really confusing. The problem is because the photocell uses a different wiring standard than is generally expected.
Generally, in the USA at least, DC power is wired with the positive (+) being red, and the negative (-) being black. This is how most of the wiring in your image is drawn.
But, the photocell has a black wire for (+), white for (-), and red for the output signal! Lots of room for mixup.
The blog author explained it in a comment:
Ah I can see how that could be confusing. This is 12VDC power, so the black wire from the power supply is ground (-), & the red wire from the power supply is positive (+). You’ll see that standard DC power wire color scheme matches the wiring that runs to the relay & actuator (and timer too, for that option).
The photocell is where the wire colors get confusing — although it is a 12VDC photocell, it oddly follows the typical 120/240VAC wire color convention as you described. That is why I have the text notes on those photocell wires. Other than the photocell, the rest of the wiring follows the standard 12VDC wire color convention of red (+) / black (-).