Find angle at which the boxes would begin to slide 
The angle θ increases causing the box of mass $m_2$ to slide. Calculate at what angle the box begins to slide. Friction factor = 0.25 ; $m_1 = 15 kg$ ; $m_2 = 25 kg$


I am having difficulty understand how my professor has solved this problem and hoped that someone might make it a bit clear or perhaps propose another way of solving this problem.
This is the diagram that my professor used

I drew a similar diagram however I did not have my positive x-axis pointing to the left but to the right. Also, I don't understand why he has that the normal force of box with mass $m_1$ is pointing downwards? Shouldn't all normal forces be in the opposite direction as the gravitational force $mg$?
Also, shouldn't the gravitational force acting upon the blue box be:
$$(m_1 + m_2)g$$
and thus the components of that force would be $$(m_1 + m_2)g\cosθ\qquad \text{respectively}\qquad (m_1 + m_2)g\sinθ?$$
I apologize if this seems off-topic but I would really appreciate it if someone could correct me if I am wrong.

 A: Actually the boxes need to be considered different because they are not rigid with respect to each other as they can slide over one another(that is distance a particle of box 1 can change with respect to particle of box 2).
Having considering only the box m2 , the gravitational force m2g acts on it and in fbd of m1 gravitational force on it will be m1g. As both blocks can move with different accelerations we need to make both block's indivisual fbd.(free body diagram)
A: You may have overlooked the string or wire shown in the picture that connects $m_1$ to some fixture, and whose tension opposes the friction and gravitational forces acting on $m_1$ in the negative x-direction. This restrains $m_1$ and prevents it from sliding down on $m_2$. This is discussed in my response to your third highlighted statement below. To be honest, I missed it myself on my initial reading. 

I drew a similar diagram however I did not have my positive x-axis
  pointing to the left but to the right.

That shouldn't matter.

Also I don't understand why he has that the normal force of box with
  mass $m_1$ is pointing downwards?

Your professor is showing a FBD of $m_2$. For this FBD the normal force $N_1$ that $m_1$ exerts on $m_2$ acts downward as shown. The normal force $N_1$ that $m_2$ exerts on $m_1$ is upward, as shown on my free body diagram for $m_1$ below.

Also, shouldn't the gravitational force acting upon the blue box be:
  $$(m_1 + m_2)g$$ 
and thus the components of that force would be $$(m_1 +
m_2)g\cosθ\qquad \text{respectively}\qquad (m_1 + m_2)g\sinθ?$$

The component of the gravitational forces of both boxes acting normal to the surface of the blue box is $N_{2}=(m_{1}+m_{2})gcosθ$, so the first term is correct. But $m_1$ does not exert a force on $m_2$ in the minus x direction because there is a string or wire shown in the picture that is connected to $m_1$ and to some fixture that opposes forces acting on $m_1$ in the negative x-direction. So the second term is not correct. It should be $m_{2}gsinθ$ as shown by your professor.
Refer again to my FBD of $m_1$ below. The tension in the string or wire in the positive x-direction equals the sum of the friction force and the gravitational force acting on $m_1$ in the minus x-direction. Consequently, the only force acting on $m_2$ in the minus x-direction is $m_{2}gsinθ$.
Finally, this problem has to do with finding the maximum angle before impending motion. There are no accelerations involved.
Hope this helps.

A: Normal forces never need to be along direction of gravity. They are the contact forces that act perpendicular to the surface in contact. I suggest read this Wikipedia Article.
A: 
Shouldn't all normal forces be in the opposite direction as the gravitational force $mg$?

No! This is a misunderstanding. The normal force has nothing to do with gravity. Normal forces can be present in space a gravity-free spaceship. Just think of the normal force as a "holding-up-force" whose only purpose is to prevent the material from breaking.


*

*For an apple on a table, sure, the normal force is upwards and gravity downwards. Because, the normal force holds up the apple and thus prevents the apple from breaking through the table. 

*At the top of a roller-coaster loop, when your head is downwards, the normal force is downwards. Because, the normal force prevents the roller-coaster cart from breaking through the track. (The cart is swung outwards in the loop due to the centrifugal effect, so the normal force pulls inwards to keep the cart in place.)

*As a third example, think of you pushing on a wall. The normal force is now horizontal, straight back into your hands. Because, the normal force prevents your hands from breaking through the wall.


In your scenario in the question, you can also see how the normal forces are not opposite to gravity. They are instead tilted at an angle.


*

*The top box feels an upwards-tilted normal force, because that normal force holds it up and prevents it from breaking through the bottom box.

*The bottom box feels its own normal force from the bottom and upwards-tilted of the same reason. The bottom box also feels a normal force from the top, because the top box is pushing downwards on the bottom box. The downwards normal force on the bottom box, is simply representing the top box's downwards pull.



Also, shouldn't the gravitational force acting upon the blue box be: $(m1+m2)g$

Nope, because the gravitational force is the weight. And the bottom, blue box doesn't weigh $(m1+m2)g$, it only weighs $m_2g$. If you removed the top box and lifted the bottom box, you would only have to lift a weight of $m_2g$. That is the only gravitational force associated with this box along. Any other gravitational forces on other boxes and other things, do not act on the box you chose to look at. The gravitational force in the top box causes that top box to push on the bottom box, so indirectly it does influence the bottom box. But you have already included that push via the downwards normal force. 
Only consider direct forces, not indirect ones.
In general, if you choose to look at just one part of a whole system - if you choose to look at only the bottom box - then you ignore what happens to all other parts of the system, and you only care about the direct influence from those other parts. The top box pushes on the bottom box, yes, and therefor you include that pushing force (the downwards normal force). That push exists because of the gravitational force in the top box, but that is not relevant when you only look at the bottom box.
