To formalize this properly, we need to adopt the viewpoint of differential geometry. So as to not be overly general, consider our system to be an ideal gas. We will call the space of thermodynamic states $\mathcal M$, and (for the moment) we will choose coordinates $(S,V)$ for $\mathcal M$ where we interpret $S$ and $V$ as the entropy and volume of a state.
A state quantity is a function $f:\mathcal M\rightarrow \mathbb R$. Examples of state quantities include the volume function, $\mathcal V: (S,V) \mapsto V$ and the internal energy function $E:(S,V)\mapsto E(S,V)$ (which is somewhat tedious to write out).
A tangent vector to $\mathcal M$ can be thought of as a directional derivative, which roughly eats a state function $f$ and spits out the rate at which $f$ changes along some direction. In coordinate form, a tangent vector $\vec X$ acts on $f$ as follows:
$$\vec X(f) = X^1 \frac{\partial f}{\partial S} + X^2 \frac{\partial f}{\partial V}$$
where $X^1$ and $X^2$ are called the components of $\vec X$ (in this basis). Accordingly, we can represent $\vec X$ all by itself as
$$\vec X = X^1 \frac{\partial}{\partial S} + X^2 \frac{\partial }{\partial V}$$
In this sense, the partial derivatives $\frac{\partial}{\partial S}$ and $\frac{\partial}{\partial V}$ are the basis vectors which span this space.
A one-form is an object which eats a tangent vector and spits out a number. Given a choice of basis $\left\{\frac{\partial}{\partial S},\frac{\partial}{\partial V}\right\}$ for the tangent vectors, we can define a corresponding basis $\{\mathrm dS,\mathrm dV\}$ for the one-forms which has the property that
$$\mathrm dS\left(\frac{\partial}{\partial S}\right) = 1 \qquad \mathrm dS\left(\frac{\partial}{\partial V}\right) = 0$$
$$\mathrm dV\left(\frac{\partial}{\partial S}\right) = 0 \qquad \mathrm dV\left(\frac{\partial}{\partial V}\right) = 1$$
and so any one-form $\omega$ can be expressed in component form as
$$\omega = \omega_1 \mathrm dS + \omega_2 \mathrm dV$$
What does it mean that $\delta Q$ and $\delta W$ are not exact one-forms?
Given any state function $f$, we can define a one-form $\mathrm df$ which eats a tangent vector $\vec X$ and spits out $\mathrm df(\vec X):= \vec X(f)$. In component notation, $\mathrm df$ takes the form
$$\mathrm df = \frac{\partial f}{\partial S} \mathrm dS + \frac{\partial f}{\partial V}\mathrm dV$$
We call $\mathrm df$ the gradient of $f$. Any one-form which is the gradient of a state function is called exact. It's reasonable to ask whether all one-forms are exact, and the answer is no. As an example, consider the one-form
$$\omega = V \mathrm dS$$
Comparing this with the general form for a gradient $\mathrm df$, we see that $\frac{\partial f}{\partial S}=V \implies f = SV + g(V)$ where $g$ is some function of $V$ alone. But we must also have that $\frac{\partial f}{\partial V} = S + g'(V) = 0$, which implies that $g'(V)=-S$. But since $g$ (and by extension, $g'$) is not a function of $S$, this cannot be satisfied. We conclude that there is no function $f$ such that $\omega = \mathrm df$, which means that this $\omega$ is not exact.
The importance of this is as follows. A quasistatic process can be represented by a continuous curve $\gamma:[0,1]\rightarrow \mathcal M$. A one-form $\omega$ can be integrated along $\gamma$ as follows:
$$\int_{\gamma} \omega := \int_0^1 \omega\big(\gamma'(t)\big) \mathrm dt$$
If $\omega = \mathrm df$ is exact, then the corresponding integral is simply the change in the state function $f$ between the endpoints of the process, because
$$\int_\gamma \mathrm df = f\big(\gamma(1)\big) - f\big(\gamma(0)\big)$$
On the other hand, if $\omega$ is not exact, then the corresponding integral is path-dependent. This is the case for $\delta Q$ and $\delta W$, which cannot be written as the gradients of some state functions $Q$ and $W$. This means that the total heat delivered to the system and the total work done by the system - given by $\int_\gamma \delta Q$ and $\int_\gamma \delta W$ respectively - are process-dependent quantities. In thermodynamic language, knowing which two states are connected by a process is not enough to know how much heat was added (or work was done) to get from one to the other.
How is $\mathrm dE$ a one-form while $\delta Q$ and $\delta W$ are not exact one-forms?
Consider the function $f = SV$. From the definition above, $\mathrm df = V\mathrm dS + S\mathrm dV$. If we define $A = V\mathrm dS$ and $B = S \mathrm dV$, we observe that neither $A$ nor $B$ are exact one-forms, but their sum is.
The first law of thermodynamics can be understood as the physical statement that during an infinitesimal thermodynamic process, the change in the state function $E$ can be attributed to some small amount of heat $\delta Q$ and some small amount of work $\delta W$. The latter two quantities are not exact one-forms because there is no "heat function" or "work function" which are well-defined for a given thermodynamic state; nevertheless, the difference $\delta Q - \delta W = \mathrm dE$ is exact.
The fact that no such functions exist can be understood by the following example. I can take an ideal gas and adiabatically compress it to a smaller volume, thereby raising its temperature. I have added no heat in this process. On the other hand, I could take the same gas, isothermally compress it to the smaller volume, and then add heat to raise its temperature. During this process I have added heat.
In both examples, I started and ended with the same thermodynamic state, but I added heat during the second process and no heat during the first. If there were a state function $Q$, then it would increase during the second process but not during the first - but both processes end in the same state! Therefore, heat is a quantity which can be associated to a process but not to a state.
