Forces involved in this circular motion system 
As seen in the image, this system involves a small mass revolving attached to a string which is anchored with a larges mass. However, investigating this made me wonder how this system is in equilibrium vertically. The two forces acting on the small mass are the centripetal force as well as its weight force but this small mass maintains a constant horizontal circular path. So is there a third force acting on the small mass to counteract the weight force?
My second question which I think might assist with understanding my first question is how do we know that the force weight of the larger mass equals the centripetal force? Aren't these two forces acting in perpendicular directions (Vertical and horizontal)?
 A: For equilibrium, the horizontal string has really to be at an angle $\theta$ as shown in the diagram.
Then resolving forces vertically on the large mass, $T=Mg$

the centripetal force is from $T\cos\theta$
and the small mass is supported by the vertical component of the tension $T\sin\theta$
A: If there was a table, the normal force would provide a force to balance the weight force on the small mass, and the string could be horizontal. The tension force would be the centripetal force (the radial component of the net force).
Without the table, the string cannot be horizontal... since it must provide an upward vertical component to balance the weight force on the small mass.
The plane of the circular path is below the upper end of the plastic tube.
The horizontal component of the tension would be the centripetal force (the radial component of the net force).
For your second question, it is likely assumed that the string is ideal (zero mass and doesn't stretch). So, in that case, the tension in the string is uniform. The bend at the upper end of the plastic tube is like an ideal pulley, whose role is to effectively change the direction of the force along the string.

UPDATE:
This update is in response to @SwissGnome's question in the comment.
It is inaccurate to say the "tension made in the string by the large mass".
The value of the tension is determined by Newton's Law using the forces acting on the string by ALL other objects---yes, the large mass, but also the small mass(!)
and negligibly gravity (since the string is idealized as massless)
and negligibly the corner of the tube (since that is likely idealized as smooth and frictionless... like an ideal pulley which changes the direction of the tension (since the tension in an taut ideal string is tangent to the string).
In this idealized case, the value of the tension in the string is the same at all points in the string.
At any point along the string, cut it there and
insert a tiny spring scale. The spring scale will read the same value of the tension at any point.
... yes, lots of assumptions that go into the Free Body Diagram (like the one drawn by @JohnHunter)].

The above description of the tension force applies in the Atwood Machine ( https://en.wikipedia.org/wiki/Atwood_machine ).

*

*Do you have an issue there with tension being the same on both sides?

In fact, your system is a variant of the Atwood Machine.
In your system, if the orbiting mass is sufficiently heavy or has a large angular speed and radius, the hanging mass will accelerate upward [if the clip is removed].
