Suppose we have a surface which has an explicit function $ z(x,y)$ then we can write the equation of surface around some point $ (x_o,y_o)$ as:
$$ \Delta z = \frac{\partial z}{\partial x}_{y} \Delta x + \frac{ \partial z}{\partial y}_{x} \Delta y$$
The geometrical picture of this is as follows:
Consider the $ z-x$ plane , in it we have a cross-section of the surface for a fixed $y$ value of $y_o$, for this curve we can write the change in height as we move $\Delta x$ as $ (\frac{ \partial z}{\partial x})_y \Delta x$ and similarly we can argue for the idea behind addition second term in the sum by considering the $ z-y$ plane.
Now, from my understanding if we have an inexact differential then it is an expression of form:
$$ f(x,y) = A dx + B dy$$
Then this can't really be considered as a differential because we can't find a surface given by an explicit function $z$ for which :
$$ (\frac{\partial z}{\partial x})_y = A$$
and,
$$ (\frac{ \partial z}{\partial y})_x = B$$
Now, we can figure out if a differential is exact or inexact by considering the mixed partial derivatives:
$$ \frac{ \partial^2 z}{ \partial y \partial x} = \frac{ \partial^2 z}{ \partial x \partial y}$$
If the above equality holds then it is exact and otherwise it isn't. An easy way that I got to think of this is by thinking of the differentials as the one forms of a vector field. Now, the vector field would only have a potential function if the differential is exact and this condition is equivalent to the vector field having zero curl.
Now, what I don't understand is how the above idea above idea of curl, vector fields etc relate to the original idea of approximating the surface? What exactly is the nature of a surface given by an inexact differential, I mean I know a surface corresponding to it doesn't exist but what if we just 'welded' together all the approximation planes at different points $x$ and $y$ someway?
In general, we want to look for an integrating factor $f$ so that the differential $1$-form $\omega = A\,dx + B\,dy$ becomes exact once you multiply by (the nowhere-zero function) $f$. This is a standard notion in beginning differential equations courses. In the plane, whenever $\omega$ is nowhere-vanishing there is always (at least locally) an integrating factor. In higher dimensions, this is not the case.
However, once you learn differential forms, you will find that the necessary (and locally sufficient) condition is to have $$\omega\wedge d\omega = 0.$$ This is a simple case of the Frobenius integrability theorem. If this condition fails, there are no integral manifolds at all, so you cannot "weld" in any meaningful way.
REMARK: In dimension $3$, I can restate this criterion for you without differential forms. If your differential $\omega = A\,dx+B\,dy+C\,dz$, define a vector field $\vec F = (A,B,C)$. Then the integrability condition becomes $$\vec F\cdot\text{curl}\,\vec F = 0.$$
But your question is far more specific. You're not asking for integral curves of $\omega$. You're asking explicitly for surfaces in $\Bbb R^3$ on which we have $dz=\omega$. So, we're asking to integrate instead the differential equation $\eta = dz - \omega = 0$. The same integrability criterion I gave earlier applies to $\eta$. We want $d\eta \wedge\eta = 0$. Here you have $\omega = A\,dx + B\,dy$ where $A$ and $B$ are functions of $x$ and $y$ only. So $d\eta = d(dz)-d\omega = -d\omega$ and $d\eta\wedge\eta = -d\omega\wedge (dz-\omega) = -d\omega\wedge dz$ (because $d\omega \wedge\omega = 0$ automatically, being a $3$-form in $\Bbb R^2$). But $d\omega\wedge dz = 0$ if and only if $d\omega = 0$, so there is no hope for integrability unless we started with a closed form.
There is a standard "physics-y" argument here. If you try to make a surface by integrating along paths $\gamma$ starting at $(x_0,y_0)$ and going to $(x,y)$, namely by setting $$z = \int_{\gamma} \omega,$$ you find that two different paths $\gamma$ and $\gamma'$ (from the same base point to the same point) result in the same integral precisely when $$\int_\gamma \omega = \int_{\gamma'} \omega.$$ If you let $\Gamma$ be the region in the plane bounded by $\gamma - \gamma'$, then Green's Theorem tells you that $$\int_\gamma \omega - \int_{\gamma'} \omega = \int_\Gamma d\omega,$$ which will be nonzero (in general) for a non-exact differential. So you cannot build a well-defined surface.