It wouldn't be easy.
You have 9 LED elements on the strip, each one has a forward voltage of around 2.5-3v, so in total a voltage of at least around 22-24v is required for them to turn on, as you found out.
Once they're lit up, you need to limit the current flowing through them, otherwise they'll simply take as much power from battery or whatever you connect the strip to, and the LEDs will blow up. That's the purpose of those 3 resistors - smaller resistance means more current flows through the LEDs and they'll be brighter, higher resistance means less current, so the LEDs will be less bright.
Now there's usually just one resistor for a series of LEDs but you have three at the start of the strip, which can mean several things :
1. the designer may have connected the three in parallel so that they'll act as a single larger resistor with smaller resistance and higher resistance to temperature (the three of them will heat less)
2. each LED in the strip may actually be formed out of three smaller leds under that yellow cover (phosphorus), so in reality you actually have three individual strips of 9 smaller LEDs.
You can put a multimeter in continuity mode and test to see if the ends of the resistors are connected together - this way you'll know if the resistors are in parallel (like I say in point 1). If there's no continuity between a couple of the resistors you know you're looking at case 2, three separate strips.
To make your strip work with 12v , you'd have to basically interrupt the connection between leds after each 3 leds. After each set of three leds, you'd have to connect the end of the led to the negative trace, you can see it if you look carefully, the trace where you solder the black wire goes all the way to the end of the board and connects to the last led.
At the start of the 2nd set of 3 leds and the 3rd set of 3 leds, you have to bring power, the red wire, and you'd also have to install three resistors between power and first led from the series, to limit the current to those 2 new strips of 3 leds.