What could be the purpose of the diode at emitter for this transistor buffer?
To understand a new circuit means to see well-known circuit concepts and building blocks in the new circuit solution. Let's apply this technique to the present circuit.

Structure. This is a digital circuit - a voltage level converter, operating as a switch with two states. The input voltage of 0...3.3 V is converted to output voltage of 0...12 V by two cascaded (connected one after another) common-emitter stages. They are implemented by complementary bipolar transistors (n-p-n Q2 and p-n-p Q1).



Concept 1. The first concept that we see here is: To drive bipolar transistors by voltage sources with V > 0.7 V, we have to insert base resistors. There are a few explanations of this need: The base resistor absorbs the voltage difference VIN - 0.7 V, limits the base current, converts the input voltage to current, enlarges the input voltage range up to extremely high values, etc. So R13 enlarges Q2 maximum base voltage to 3.3 V… and R12 enlarges the Q1 base voltage to 12 V. The first voltage is obvious - this is the output voltage of the microcontroller port. The second is the 12 V supply voltage. It is applied to Q1 base through R12 when Q2 is on.

Concept 2. The next concept is to introduce threshold in the input of switching circuits. Q2 has the inherent Si base-emitter voltage threshold of 0.7 V but they decided to increase it. The simple solution is to add another voltage drop of 0.7 V by inserting another (forward biased) diode in series to Q2 base-emitter junction. We can do it in two ways - before the base and after the emitter. The difference is in the current flowing through the diode. In this case it would be beta times smaller than in the picture… and the voltage drop across the diode would be smaller than 0.7 V (look at the diode IV curve). That is why they inserted the diode in the emitter where the big emitter current flows. Indeed, this "lifts" also Q2 collector voltage when saturated but this is not important in this configuration.

Concept 3. Another very important concept that we can see, is: Never leave the transistor base “floating” (unconnected to ground). The reason is that, in this case, it is vulnerable to noises, leakages and static electricity. Hence the role of R11 and R14. 

In particular, the role of R11 is to shunt Q1 base-emitter junction when Q2 is off while the role of R14 is to protect Q2 base when the circuit input is left floating (the microcontroller is not connected  to it).  

R11-R12. Only, R11 and R12 constitute the basic circuit building block voltage divider. It has to be calculated so that when Q2 is on, the voltage drop across R11 to be more than 0.7 V - VR11 = 12.R11/(R11 + R12) > 0.7 V. With some reserve, this means R11 > 1 k. When Q2 is off, the junction is high resistive and R11 can be high (hundreds of koms, typically 100 k).

So the value of R11 is not so important; it can vary widely since it is connected in parallel to Q1 base-emitter junction... and we can assume the range of R11 is 1 ÷ 100 k. The role of R11 as a "pull-up" resistor is minor when Q2 is on; then Q1 base-emitter junction "pulls up" R11-R12 midpoint. R11 really "pulls up" it when Q2 is off.

R13-R14. Similarly, R13 and R14 constitute another voltage divider. It can be calculated in a similar way so that when VIN = 3.3 V, its output voltage (across R14) to be more than 1.4 V...