We know what is the interpretation of a total differential, ex.: $$df=\frac{\partial f}{\partial x} dx+ \frac{\partial f}{\partial y} dy$$
But what is the interpretation of a 1-form and its exterior derivative equal to 0, ex.:
$$g=xdx+ydy$$ $$dg=0 \:dx \wedge dy$$
Is there any link connecting both concepts?
You were actually looking at a closed form $g$, i.e. $dg=0$. $df$ is called an exact form, which is automatically closed since $d^2=0$. But not all closed forms are exact, i.e. equals to the differential of a form with one order lower. On a simply connected domain, the two types coincide. For further story, you should check deRahm cohomology. Take your example, $g=xdx+ydy, dg=0$ (assuming the domain being all of $\mathbb R^2$ and therefore simply connected), it means $g$ is exact, i.e. $g=df$. Here $f(x,y)$ is obviously given by $x^2/2+y^2/2+\text{const.}$.
The interpretation of a 1-form $dg=0$ is obvious: $g$ is a constant function. Or when $dg(x,y)=0$ at a particular point $(x,y)$, it is said to be a stationary point. It has much physical meaning as you probably already know. On a $n$-dimensional domain $D$ with boundary $\partial D$, Stokes theorem says: $$ \int_{\partial D}g=\int_Ddg $$
where $g$ is an $(n-1)$-form. If $dg=0$, $\int_{\partial D}g=0$. This equation probably has a dozen physical meanings, such as being divergenceless as in the case of divergence theorem and being curless as in the case of curl theorem. The details involve sharp/flat/hodge/exterior differentiation operators.