The relation between motivic cohomology and cohomology of the Milnor K-theory sheaf is discussed in

Motivic cohomology and cohomology of Milnor K-theory sheaf

There is a natural comparison morphism ${\rm CH}^2(S,1)={\rm H}^3(S,\mathbb{Z}(2))\to {\rm H}^1_{\rm Zar}(S,\mathbf{K}^{\rm M}_2)$ coming from the edge maps of the coniveau spectral sequence for motivic cohomology. Most of the relevant input is contained in Bloch's paper "Algebraic cycles and K-theory", in particular the relevant spectral sequence is the local-to-global spectral sequence (iv) on page 269 of Bloch's paper.

In the case of a smooth surface $S$ over a field, this spectral sequence for weight $n=2$ degenerates at the $E_2$-term because it only has entries $E_2^{p,q}={\rm H}^p_{\rm Zar}(S, \mathcal{H}^q(n))$ in the column $p=0$ and the row $q=2$. Therefore, the edge map ${\rm CH}^2(S,1)\to {\rm H}^1_{\rm Zar}(S,\mathbf{K}^{\rm M}_2)$ is an isomorphism. This is the isomorphism between the descriptions (1) and (2) in the question.

Let's make the edge map isomorphism more explicit. The corresponding filtration on motivic cohomology/higher Chow groups ${\rm CH}^2(S,1)$ is given by codimension of support. The relevant cycles are supported on curves, since ${\rm CH}^2(F,1)$ is trivial for a field $F$ (like the function field of $S$). In particular, for any cycle in ${\rm CH}^2(S,1)$ there exists a curve $C\subset S$ such that the restriction ${\rm CH}^2(S,1)\to {\rm CH}^2(S\setminus C,1)$ maps the cycle to zero. The edge map of the spectral sequence maps ${\rm CH}^2(S,1)$ to the quotient modulo those cycles supported on points, in particular, we are free to remove finitely many points from $S$. So we can assume that the curve in $S$ is smooth, and use localization for higher Chow groups:
$$
{\rm CH}^1(C,1)\to {\rm CH}^2(S,1)\to {\rm CH}^2(S\setminus C,^1).
$$
Any cycle in ${\rm CH}^2(S,1)$ that vanishes upon restriction to $S\setminus C$ must already be a cycle in the image of $C\times\Delta^1$. Again, we can remove more points of the curve, whence our cycle is simply given by a collection of points in $\Delta^1$ over the function field of the curve. We know that this corresponds exactly to the units of the function field. What we have written down is an explicit mapping from ${\rm CH}^2(S,1)$ modulo the cycles supported on points to $\bigoplus_{x\in S^{(1)}}\kappa(x)^\times$, and this mapping induces the edge map of the coniveau spectral sequence.

Concerning the obvious idea with the graph: I think this is a way of writing some sort of inverse of the above map. However, the map goes the wrong way, and we should be aware that this map being well-defined requires that we know a priori that the edge map is an isomorphism. The problem with the poles of the rational function also becomes clear then: the edge map is obtained by taking the quotient modulo cycles supported on points, so we are free to remove an arbitrary finite number of points from the surface. Doing this allows to remove all the poles of the rational function on $C$, making the graph of the rational function a cycle on $S\setminus \{p_1,\dots,p_n\}$. The vanishing of $\mathcal{H}^{-1}(\mathbb{Z}(0))$ which implies that the edge map is an isomorphism also implies that the restriction map ${\rm CH}^2(S,1)\to {\rm CH}^2(S\setminus \{p_1,\dots,p_n\})$ is an isomorphism.

Concerning the last question. It's not clear to me that the map is at all well-defined, the nature of the relations in ${\rm CH}^2(S,1)$ and ${\rm CH}^2(S\times\Delta^1)$ seems rather different. Moreover, assuming that the map is actually well-defined, it should be the zero map by the following argument. Homotopy invariance implies ${\rm CH}^2(S\times\Delta^1)\cong {\rm CH}^2(S)$. The isomorphism is induced by pullback along the projection, taking a zero cycle $\sum n_i[Z_i]$ to $\sum n_i[Z_i\times \mathbb{A}^1]\subset S\times\mathbb{A}^1$. These cycles are supported on points, and hence do not contribute to ${\rm CH}^2(S,1)$. For a concrete example, ${\rm CH}^2(\mathbb{P}^2,1)\cong F^\times$ by the projective bundle formula, where $F$ is the base field. Over many fields, the map ${\rm CH}^2(\mathbb{P}^2,1)\to {\rm CH}^2(\mathbb{P}^2\times\Delta^1)\cong \mathbb{Z}$ has no choice but to be the zero map.