The following form
$$M(x,y)dx + N(x,y)dy$$
can be derived from the total differential of a bivariate function $u(x,y)$, which is
$$du = u_x(x,y)dx + u_y(x,y)dy.$$
Therefore, $du = M(x,y)dx + N(x,y)dy$.
This is perfectly understandable so far.
Then we somehow set $u(x,y) = c$, which makes the differential of $u$ equal to $0$.
I'm confused about this. If $u(x,y) = c$, then $y$ is no longer an independent variable. The value of $y$ is now dependent on $x$.
Simply put, $u(x,y) = c$ is equal to $u(x,y(x)) = c$. It's really just an implicit form of a single-variable function.
So how is $du$ even defined in this case? Why is $du$ equal to $0$ when there's no tangent plane to approximate the value of $\Delta u$ in the first place?
I've always understood the total differential $$du = M(x,y)\,dx + N(x,y)\,dy$$ in this context to be notational shorthand for $$\frac{du}{dt} = M(x(t),y(t))\,\frac{dx}{dt} + N(x(t),y(t))\, \frac{dy}{dt}.$$ for any differentiable curve $t \mapsto (x(t),y(t))$.
Along this vein, $du=0$ on $u(x,y) = c$ means that for any differentiable curve $t \mapsto (x(t), y(t))$ on $u(x,y)=c$, we have $\frac{d}{dt}u(x(t),y(t)) = 0.$
At this point, it should be obvious. If $t \mapsto (x(t),y(t))$ is a curve on $u(x,y)=c$, then for all $t$, $u(x(t),y(t))=c$, and the derivative of a constant is zero.