Heat and work are not the state functions of the system. Why?

Suppose you have a thermodynamic system in a state $A$. In this state it has a certain amount of internal energy, $U_A$, because internal energy is a state variable. You can determine the internal energy by knowing only the state.

Now suppose the system undergoes some process - you don't know (or care) what - that leaves it in state $B$. Again, you can determine its internal energy by knowing only the state, and that energy will be $U_B$.

Clearly, to get from state $A$ to state $B$, the system had to gain a net amount of energy $\Delta U = U_B - U_A$. That fact is true no matter what the process was. Hopefully that's all clear so far.

But hopefully it's also clear that how that energy was transferred to the system depends on the process. There are many - or, well, at least several - different ways by which energy can be transferred in or out of a system, such as electromagnetic radiation, sound, gravity, or physically pushing on something. Different processes will use different methods, or different combinations of methods (because it's possible for a process to transfer some energy by EM radiation and some by sound, for example).

We've broadly grouped these methods of transferring energy into two categories, heat and work. Generally, the methods which involve the system pushing its environment around (or vice-versa) count as work, while others count as heat. (This should make sense because you need force and a change in position to have work.) So depending on the process by which the system gets from $A$ to $B$, the amount of energy transferred by heat methods and the amount of energy transferred by work methods can vary. Any basic thermodynamics textbook will give several examples to show how the distribution of energy transfer between heat and work depends on the process.

Of course, the total amount of energy transferred through all methods is always the same: it always has to be $\Delta U$. That's just energy conservation.

Think of it like this. When you have an object of mass $m$ which is held a height $h$ above some reference point, you think of it as having potential energy (considering only gravitational interactions) $U= m g h$, and gravity will exert an amount of work $W_g = m g h$ on the object. When you drop the object, it shall fall towards the ground, towards “equilibrium”, so to speak. You do not speak of the amount of “work” that the mass has when at its original height, nor of the amount of “work” lost, but of its energy (relative to a reference point) at any given state, $U$. Moreover, we say that this potential energy is a state function because it depends only on the initial and final heights of the mass in question.

In the same way, one does not concern his or her self with the amount of “heat” that an object has, since it is merely a term used to denote the amount of transferred energy between systems as they move in and out of equilibria. We speak of thermal energy, internal energy, free energies and such that are state functions of the system - in exactly the same way that the gravitational potential energy $U$ was in the mechanical analogue to this thermodynamical case. In the same way, we say that the thermal energy of the system is a state function insofar that it generally depends (more or less) on the initial and final temperatures and thermodynamic quantities of the mass in question.

Edit: I’ve reread your question and I want to make another point to clear things up. Yes, indeed, different paths can result in different amounts of heat transfer - the first law of thermodynamics states:

$\delta E = Q + W$, wherein $Q$ is the amount of heat flow into the system, $W$ is the work done onto the system, and $\delta E$ is the total state internal energy change of the system. One can see that one can input say, 100 J of heat and do no work on a system to result in a net change $\delta E$ of 100 J, and in the same way, one can divide that $100 J$ amongst $W$ and $Q$ to get the same effect.

The intuition is as follows. Imagine you have a jar of gas. You can increase the temperature (and so impart a positive $\delta E$) by adding $100 J$ of heat, or you may compress it by doing $100 J$ of work to gain the same effect. I hope that clears things up!

You and a friend are in front of a mountain. You both are at 100m elevation. Therefore you are in the same state. Now you walk up to the top, your friend takes the bus. Then you both compare how much energy you spent on your ways. You spent a lot more than he did. A third person comes and shows you the amount of energy he needed to go up by bike. 3 different levels of work done. But still you are all at the same state, for example 1100m elevation.

A function works like f(x) = y. If someone tells you how much energy he spent, can you tell him exactly, how high he got? If not, "energy he spent" is not a function of the state.