Summary
As proposed, your LEDs MAY die a bright and early life.
It's easy to help them to live long and prosper.
It is easy to drive LEDs properly.
Failing to do so can lead to very short lifetimes and uneven illumination.
LEDs should always be operated with either a constant current supply or a supply which is close enough to constant current for the purposes. Operating them from constant voltage directly invite substantial learning experiences which you might rather not have.
A small decrease in LED_Vf at a given voltage can lead to a very substantial change in LED current. If you parallel 4 LED strings and do not make any attempt to equalise currents then you can be almost certain that string currents will not be equal.
One easy way to set maximum current in a string is to use an LM317 and one resistor. This requires about 3.5 V of voltage drop to function so may or may not suit in your case. The following circuit is from fig 19 in the LM317 datasheet.
As shown, R1 sets maximum current - here for 750 mA R = 1.25/0.750 = 1.7 ohms. You could use the standard E12 value of 1.8 ohms or a lower value eg 1.5 ohms to cause limiting once imbalance got too high (I = V/R = 1.25/1.5 = 830 mA).
Without some form of per string limiting:
Current imbalance of up to about 2:1 per string MAY occur.
If any one LED blows total average string current rises to 3A/3 = 1A. If you again got 2:1 imbalance then currents may be eg 750 mA, 750 mA, 1.5A in the 3 strings. If your LEDs are rated at a true 3W and the Vf you have stated occurs in practice then Imax = P/V = 3/2.2 = 1.36A SO the LEDs would not be vastly over-rated at 1.5A. Still not a good idea though.
If you have lots of "headroom" you can use more voltage and a series R to drop voltage and stabilise current. An LM317 will give a much more stable result. The LEd shown here can be one LED or a string. Drop aross the LM317 circuit is 1.25V across the resistor plus the dropout voltage of about 2 Volts (graph on page 6 of datasheet).
LM317 or other current source power dissipation
Each LED string is specd as drawing 750 mA.
The LM317 has a minimum dropout of 2V and dropout will be as high as required to maintain constant current.
At 2V that's 1.5 Watt (2V x 0.75A) and at say 5V it's about 4W. A 10 degrees C/Watt heatsink will handle that OK. If you need higher than that there is something wrong with the design. The higher dissipation MAY be required if one string of 4 goes O/C but that's a fault condition. The heatsink may be thermally shared with all 4 x LM317 as they will only go into high voltage drop when something goes wrong and drop will vary depending on LED Vfs.
The LM317 has thermal shutdown and will gracefully turn off as/if required due to overload.
Note that the ability of the LM317 to "float" is part of what allows it to be used as a constant current source. It obtains it's internal power supply from the drop across the regulator. Other regulators may have a low minimum dropout BUT rely on a higher value from Vin to ground terminal to power them.
As soon as the string goes into constant current mode the regulator (regardless of dropout voltage) will be required to drop any "excess" voltage so even a low dropout regulator will be very little better off when it is called on "in anger".
I have used LM317's as constant current drivers for LED strings on numerous occasions (usually for LED testing) and find them very useful for this purpose, subject to proper thermal design. For currents above 1A (1.5A in some versions) an LM350 may be used (up to 3A afair).
Your "LED Driver" is most likely bad
The driver is clearly out of spec and mostly likely internally damaged. From you photos it isn't clear how the rest of your lights and power sources are connected, but you may have made an error here (I've even seen people connect the AC power line to the DC output side of the supply).
You have a current source, not a voltage source
If you look at the label you will notice that current is specified precisely (600mA) and voltage is specified as a range (16v-28v). You will also notice that the drawings on the label show a single current loop and specify 7 3-Watt LED's in series.
That you provided this equation indicates that you are confused about the difference between the two types of sources:
$$R=\frac{V_{Driver}-V_{LED} \times 7}{I_{LED}} = \frac{35\ V-3.7\ V \times 7}{0.7\ A} = 13\ \Omega\ $$
In your equation, you cannot know the value of $$V_{Driver}$$ as it is determined by the network (it's somewhere between 16 and 28 volts if an acceptable load is attached). Only the current value is constant in the normal operating condition.
Some background
A voltage source presents a single output voltage, no matter what you connect to it. To make that a true statement it will output any current the load wants up until it is incapable of outputting any more (current-limit or failure). Most people are familiar with this behavior as it is intuitive and commonly encountered.
A current source will attempt to output a constant current no matter the load attached to it. It will do this by adjusting the output voltage until it either can raise it no further (limit of it's upward adjustment range) or lower it no further (it will produce insufficient voltage to operate itself and shutdown).
This works via Ohm's law (V = I R) such that increasing the voltage will increase the current flowing and decreasing the voltage will decrease the current. The system is active and senses it's output current (while adjusting its output voltage) until the output current equals the number printed on the label (in this case 600mA).
If nothing (or too little load) is attached, it will output it's maximum voltage as it keeps trying to increase voltage to get increased current... and vice versa if too much load is attached.
Driving LED's
If your LED's are connected as parallel strings of series lights, they will need to be driven by a voltage source. This configuration is cheaper to design and build, but more difficult to install and subject to greater line losses as the LED strings get bigger (since you need to bring the full voltage of the power supply all the way to the end of the line).
If your LED's are connected in series (one to the next), then the same current that drives one light will drive the next. This configuration is used in most higher-end architectural lighting. You use a current source to ensure that no matter how many lights are on the string, the current output remains the same. The advantages are that you can easily add lights to existing strings without worry. The fact that current flows through all lights ensures that voltage losses in the line are minimal (most efficient power distribution). And, LED light output is proportional to current so a current driven approach best ensures uniformity of light output.
There are some limitations however. Current drivers are more expensive as they are produced in lower quantities and have to be matched to the exact fixtures being used. The fixtures must all be the same so that they have the same light-to-current relationship and will not be too dim (or burn up) under the constant current value applied. Series wiring of light fixtures is inconvenient in some installations.
Best Answer
as you mention these boards are from Headlamps, with LED board did you find another board inside the headlamp?. because the board you are missing is LDM (LED Drive Modules). This is a self-contained power supply, that accommodates the LED’s powers. The modules have multiple channels to provide varying currents to different LED’s. These modules are used to power LED’s which are used to customize to accommodate the LED’s it powers. if you have that board, then you just connect the board with the LDM and give power to the LDM between 9V to 16V as it is headlamp it can take up to 6Amp of current approx, and check the results. if you didn't find the board don't worry, but you can still turn on these LEDs but not recommended. use the power supply in CC mode to drive the LED board.in my case, what I have done, I'll explain in brief, I have APlab 30V10Amp variable power supply, connected the LED board to power supply and I keep the current knob to 0 and start increasing voltage knob to around 6 volts as I increase the voltage PS goes into the CC mode then I started increasing the current the LED start glowing. it worked for me maybe it will work with you. note:- do not run the LED for more than 10 sec. they are high power LED and without proper heat management the led may damage.