So, percent yield is always the same calculation, no matter the reaction. Look at your starting materials, and figure out the moles of each. Whatever has the least moles is your limiting reagent. Now calculate the moles of product you get. Divide moles of product by moles of limiting starting material to get % yield - the % of starting material you managed to convert to product. Do this independently for every step of your synthesis. To get a total overall yield for the whole synthesis, simply multiply the % yields for each step together. If you get 50% yield on the first step, and 50% yield on the second step, that's 25% overall yield.

As for the mechanism: In the first step, you have the acyl chloride, which is a great electrophile, and the dimethyl pyrrole rings, which are decent nucleophiles. You have electron density at three positions on the pyrrole: the nitrogen, and the two H-bearing carbons via resonance. Of these, the carbons should have more electron density, and the one closer to the nitrogen is less sterically hindered by the methyl groups. So it attacks the acyl chloride, knocking off the chlorine. That leaves you with a ketone, still a decent electrophile, so the second pyrrole attacks. Follow with dehydration. You should have one nitrogen protonated, one not.

Boron trifluoride etherate is in equilibrium with boron trifluoride plus diethyl ether. BF3 is a great lewis acid. So the BF3 accepts the electron pair from unprotonated nitrogen. Your triethyl amine then deprotonates the amine, leaving you with a formal positive charge on nitrogen, and BF3, which has a formal negative charge. What happens next is a bit harder - I suspect that TEA first grabs a proton from the protonated nitrogen, and then a fluorine atom grabs the proton from positively-charged TEA+, leaving you with HF, TEA, and N-BH2. BH2 is a good lewis acid once again, so it accepts the electrons from the other nitrogen, which is deprotonated with a negative charge. This gives you the final, charge-balanced molecule.

You should calculate the yield in respect to 2,4-dimethylpyrrole.

main raw material molecular weight will be used in yield calculation

Thank you, everyone.

I pay my deep sense of gratitude to Dr. William P Katt who took some time (out od his tight schedule) to explain the scene so gracefully.