What you want to do is potentially* doable but there are several points to check.
The halogen bulbs are probably AC operated.
The LED bulb example you gave appear to be DC operated (as would be expected) and LED strips in general will be DC operated.
LED lamps that accept AC are liable to be available but less common.
If you simply rectify AC to full wave DC you get pulsating DC as far as the LEDs are concerned. While halogen output falls as approximately V^2, the filament thermal time constant provides substantial light output across the voltage "valleys". LED strip light output falls very approximately linearly with V at first (as I approximately tracks V around operating voltage) but falls to zero as voltage decreases to much below 3V per LED. Most LED strips rated for 12V operation have 3 white LEDs in series per section so need > about 9V to operate at all.
If you add a smoothing capacitor to rectified AC it has to maintain operation across the voltage valleys. Capacitor voltage drop is ~~~= t.I/C.
So for eg 1000 uF = 1 mF = 0.001F, and t = 8 mS (half cycle at 60 Hz) and 1A (12W at 12v)
V ~= t.I/C = 0.008 x 1 / .001 = 8V = too much
10,000 uF gives about 0.8V in the above equation at 1A/ 12W = perhaps OK.
Easier is to replace the halogen supply and to operate the LEDs from a 12VDC power supply which has looked after the smoothing requirements for you. Current rating to suit the strips used.
I looked at a range of LED strips in the Shenzhen markets and found that light output and Wattage input varied widely. This is NOT due to greatly improved efficiency by some but due to some being willing to run their LEDs at higher currents. In some cases this may be because the LEds are rated at higher currents, but it may just as well be because they know that brighter strips will sell better. Even if the LEDs ARE rated at higher currents, power dissipated will increase and adequate heat sinking will be required to keep temperatures down. Most LED strips have an aluminum strip backing which acts as a heatsink. Temperature rise will not usually be enough to cause a fire hazard but cheap non-name LEDs plus high temperatures may led to short lifetimes. It may be that operating LED strips conservatively will pay lifetime dividends. Maybe a series resistor if they seem brighter than reasonable. The resistor needs to be adequately power rated and put somewhere where its temperature rise causes no problems. If a variable voltage supply is used (eg using an ebay etc buck-boost supply) the brightness can be adjusted to suit.
LED strips can be equipped with Aluminum mouldings that the strips slide into. These can provide improved heatsinking due to increased surface area - but check to see that this does happen.
You could improve heatsinking of LED strips by sticking them on an Aluminum "L" extrusion and then sticking that in place with DS tape. The extrusion can be arranged to have superior air flow, area and thus overall cooling characteristics.
What country are you in?
*pun noticed.
Aluminum = = Aluminium = = Alumium.
It appears that my code is to blame (scratches head). I loaded NeoPixel's strandtest and all 8 LED modules lit up on the WS2812 once that was loaded on the 'Duino.
I will need to experiment further taking into account all of the helpful suggestions to see if I can get my code to work with only hardware modifications.
In response to the helpful people who responded in the comments section A) Thank You!, B) I did have a filtering cap on the power rails. I didn't add it to the schematic but you can see it in my photo. You can also spot a resistor on the Data In line for the 8x WS2812. C) I will have to experiment with shorter wires and different libraries. I plugged Vcc and GND directly from the Arduino to the horizontal row the 8xWS2812 was plugged into and that didn't change any of the behavior. I plugged the Data In line from the Arduino to the vertical rail of the breadboard and that didn't seem to help either.
I will experiment further and report back.
Best Answer
Sounds like they either "burnt out", or you have a break in the wire, close to the supply. According to specification on the sellers website, for over voltage burnout you would need to exceed 14.8V.
Which for a truck, actually is not unlikely. Many trucks run parts at 24V with 2 batteries, but I'll assume you haven't made that mistake.
Lead Acid batteries have particularly variable output voltage depending on current charge level. Potentially the Alternator may also vary output voltage depending on motor speed (I know little about alternators).
Check to see yours does not sometime operate at ≥15v using a multimeter. If it does, then it is likely you have destroyed the LEDs.
You could also check this by testing if the LEDs still work if you power them with a more trustworthy power supply, Eg a Mains-DC adaptor you have checked with a multimeter -- or if you have access to a lab regulated voltage supply, that is ideal.
The other option is a break in the wire. This must be occurring at one of the ends, before any of the lights, as they are in parallel, so if the break was in the middle, some lights would still work.
Alternatively you have managed to separately break a wire in each group of 3, up to the point where there is a break in the series connections between groups.
Some notes I used while working this out:
WSFS-NW180-BK LED Light strip is made of 3528SMD LEDs
As it can be cut into subsections so long as each subsection contains 3 LEDS, this would indicate that the circuit is made of parallel groups of 3 in series, looking closely at this picture I believe there is a resistor in there, but I can't make out the numbers, so don't know the resistance.
simulate this circuit – Schematic created using CircuitLab
So since you have groups of 3 in series, you have a voltage divider situation. We can tell that \$V_{LED} < \frac{V_{supply}}{3}\$
According to the sale website the strip is rated for \$9 \le V_{supply} \le 14.8 VDC \$
According to the Datasheet for the white LED, to operate \$2.8 \le V_{LED} \le 3.6\$