The responsively number given in the table is specific to your device and is exactly what you need (but only partly - see below). There is no "single" parameter for all sensors as it changes from manufacturer to manufacturer. This is primarily determined by QE (Quantum Efficiency) both internal and external QE that is all bundled up in the one number of responsivity.
What you need is a mapping from Lux to Watts, and then the responsivity maps from watts to current.
All detectors will need a passivation layer on top of them to protect the underlying detector material (Here it's Si) so you'll have layers of SiO2 and other material on top. This is important as the External -QE is concerned with getting the light into the Si. This is explained using fresnel equations, but is best understood by the need to match the index of refraction in air (~ 1.0) to that of Si (~ 3.8), the use of AR (Anti-reflection) coatings, and the interaction of light with the passivation layers greatly affects the external QE of the sensor. Once the light gets into the sensor, internal QE is now the concerning factor. As the light penetrates the Si, it leaves a trail of E/H pairs (electron/hole) which are then swept up in E-fileds in the Si substrate. While the E/H generation is understood the E-fields are what determine which electrons/holes get collected. If you generate a E/H pair but it doesn't get collected then you lose internal QE. The electric fields are in turn created through the distribution of dopants and the applied voltages to the device.
In short, even though the Si absorption characteristics are well understood, individual diodes can vary wildly with design. The good news is that this can determined with the appropriate experimental setup. For example the QE of image sensors (say in the green) can vary between manufacturers from as low as 20% up to 98%. In teh NIR (say around 850 nm) these values diverge even more from 1% to 40%.
Radiometry is the measurement of light in quantitative units, Lux is the same curves with the human photopic response laid over top. Consider that mapping as a dimensionless attenuation factor that is dependant upon wavelength.
Ideally what you have is the illumination vs wavelength spectra, the photopic curve again vs. wavelength (which is easily found on-line) and the sensor response vs. wavelength and from those you'd calculate the amount of current flowing.
You have two deficiencies though. One is that you have not identified your illumination spectra and two, the sensor is only defined at 3 points.
A short hand way of calculating is to use the simple estimate (and it will be only an estimate) of 1 lux =\$\frac{1}{683} \frac{watts}{m^2}\$ @ 556 nm (green). Basically this is saying that if you have a green laser at \$ 1 \frac{w}{m^2} \$ then it will appear as 683 Lumens to the human eye.
You will need to understand the difference between luminance and illuminance. So this means you will need to also say what the imaging/collections system is and in particular it's F/#.
Knowing the relationship between wavelength and energy for light \$ E = \frac{hc}{\lambda}\$ where h = planck's constant, C = speed of light. Will allow you to determine the photon flux. And from that you can come up with the shot noise of the system.
Once you can provide the illuminant wavelength dependance, the collection optics f/# and various other parts I'll come back and fill in the details. Or if you want to use the pointers here to answer the question I can check out the answer for you.
First, the Lighthouse puts out a lot more optical power. Instead of a puny little 5 mW laser, they apparently use a string of lasers, and you can easily get 10 times more power from that sort of setup.
Second, you seem to have missed the basic idea about emitters/receivers entirely. If your laser produces a visible line (and it seems to do so), why on earth would you try to detect it with an IR receiver? The BPW34 data sheet www.vishay.com/docs/81521/bpw34.pdf gives it a peak sensitivity at 950 nm, but it also works at visible wavelengths, but at lower sensitivity. You're using a visible laser, so I'd expect lower sensitivity from your setup than is possible with an IR emitter of the same power. And going to a BPW FS will only make matters worse, since the FS incorporates a visible-rejection filter, so it won't see your visible laser at all. This also explains why you could not detect other lasers with an IR receiver, although there are other issues such as modulation which you need to learn about.
What you do need is a narrow bandpass filter at your laser wavelength. You can find them on eBay. If you get something, keep in mind that, for it to be really effective, you'll have to make a light-tight housing for your diode so that only light which has gone through the filter hits the diode. You really need to do something like this anyways. Try this experiment. With the laser off, put a box over your PD and take a reading. Call this your zero point. Now take the box off and take another reading, and call this your ambient level. Finally, turn on your laser and take a third reading - call it your active level. If the ambient is significantly different from zero (and it will be) this represents a laser light level which is unmeasurable, since the laser is being masked by the ambient.
An amplifier is probably a good idea, but this is clearly going to be a learning experience for you.
Best Answer
You're going to need six copies of the circuit on the breadboard. The output of each circuit connects to a different "ANALOG IN" pin on the Arduino.
There's no problem with sharing +5V and Gnd among the six circuits.
If space is at a premium, you might consider switching to the LTC1051 or LT1053, which are dual and quad versions, respectively, of the LTC1050.