The answer to question 1, which asks whether this definition recovers homotopy groups of a manifold, is yes. As in the question, we'll take $M$ to be manifold, $G_0 = \bigsqcup_i U_i$ the disjoint union of the sets of an open cover of $M$ and $G_1 = G_0 \times_M G_0 = \bigsqcup_{i,j} U_i \cap U_j$. It is easy to see that $G_2 = \bigsqcup_{i,j,k} U_i \cap U_j \cap U_k$, and generally, $G_n$ is the disjoint union of the $(n+1)$-fold intersections of the $U_i$. To prove that $B\mathcal{G}$ has the same homotopy groups as $M$, we'll show in fact that these two spaces are homotopy equivalent.

First of all there is a clear map $B \mathcal{G} \to M$, that sends any point $(x,p) \in G_n \times \Delta^n$ to just $x$. This map is well defined since (1) $G_n$ is just a disjoint union of some open subsets of $M$, so any point of $G_n$ can be thought of as a point of $M$, and (2) all the face and degeneracy maps in $\mathcal{G}$ just perform bookkeeping: they change which multiple intersection of $U_i$'s a point is regarded as being in, but don't actually change the point.

Now, we'll define a section $M \to B \mathcal{G}$. For this, take a locally finite partition of unity $\phi_i$ subordinate to the cover $\{U_i\}_i$. Given $x \in M$, it'll belong to the support of finitely many $\phi_i$, say $\phi_{i_0}, \ldots, \phi_{i_k}$ and we can send $x$ to the equivalence class of $(x, (\phi_{i_0}(x), \ldots, \phi_{i_k}(x))) \in G_k \times \Delta^k$. Here, $x$ is thought of as belonging to the $U_{i_0} \cap \cdots \cap U_{i_k}$ term of $G_k$, and we regard $\Delta^k$ as the set of points $(t_0,\ldots,t_k) \in \mathbb{R}^{k+1}$ such that $t_i \ge 0$ and $t_0 + \cdots + t_k = 1$. It's straightforward to check that the definition above does give a function $M \to B\mathcal{G}$ (for example, adding extra $\phi_{i_j}$ with $\phi_{i_j}(x)=0$, does not change the image of $x$ in $B\mathcal{G}$), and it is clear this map is a section of the map $B\mathcal{G}\to X$ above.

Finally, the composite $B\mathcal{G} \to M \to B\mathcal{G}$ is homotopic to the identity by a straight line homotopy within the $\Delta^k$'s (that is, via homotopy that is the identity on the $G_k$ coordinate and moves in a straight line segment in the $\Delta^k$ coordinate).

This argument has probably been discovered many times and is well-known, I fairly recently found out that it does appear in print at least in Segal's Classifying Spaces and Spectral Sequences, proposition 4.1. It is also true that even if $M$ is not a manifold but rather just a topological space with a cover $U_i$ possessing no subordinate partition of unity, that the homotopy groups of $M$ and $B \mathcal{G}$ are the same: these two spaces are weakly homotopy equivalent. See Dugger and Isaksen’s Hypercovers in Topology, theorem 2.1. This simplicial space $\mathcal{G}$ is called a Čech cover, and Dugger and Isaksen's paper also deals with the more general notion of a hypercover.

The answer to question 1, which asks whether this definition recovers homotopy groups of a manifold, is yes. As in the question, we'll take $M$ to be manifold, $G_0 = \bigsqcup_i U_i$ the disjoint union of the sets of an open cover of $M$ and $G_1 = G_0 \times_M G_0 = \bigsqcup_{i,j} U_i \cap U_j$. It is easy to see that $G_2 = \bigsqcup_{i,j,k} U_i \cap U_j \cap U_k$, and generally, $G_n$ is the disjoint union of the $(n+1)$-fold intersections of the $U_i$. To prove that $B\mathcal{G}$ has the same homotopy groups as $M$, we'll show in fact that these two spaces are homotopy equivalent.

First of all there is a clear map $B \mathcal{G} \to M$, that sends any point $(x,p) \in G_n \times \Delta^n$ to just $x$. This map is well defined since (1) $G_n$ is just a disjoint union of some open subsets of $M$, so any point of $G_n$ can be thought of as a point of $M$, and (2) all the face and degeneracy maps in $\mathcal{G}$ just perform bookkeeping: they change which multiple intersection of $U_i$'s a point is regarded as being in, but don't actually change the point.

Now, we'll define a section $M \to B \mathcal{G}$. For this, take a locally finite partition of unity $\phi_i$ subordinate to the cover $\{U_i\}_i$. Given $x \in M$, it'll belong to the support of finitely many $\phi_i$, say $\phi_{i_0}, \ldots, \phi_{i_k}$ and we can send $x$ to the equivalence class of $(x, (\phi_{i_0}(x), \ldots, \phi_{i_k}(x))) \in G_k \times \Delta^k$. Here, $x$ is thought of as belonging to the $U_{i_0} \cap \cdots \cap U_{i_k}$ term of $G_k$, and we regard $\Delta^k$ as the set of points $(t_0,\ldots,t_k) \in \mathbb{R}^{k+1}$ such that $t_i \ge 0$ and $t_0 + \cdots + t_k = 1$. It's straightforward to check that the definition above does give a function $M \to B\mathcal{G}$ (for example, adding extra $\phi_{i_j}$ with $\phi_{i_j}(x)=0$, does not change the image of $x$ in $B\mathcal{G}$), and it is clear this map is a section of the map $B\mathcal{G}\to X$ above.

Finally, the composite $B\mathcal{G} \to M \to B\mathcal{G}$ is homotopic to the identity by a straight line homotopy within the $\Delta^k$'s (that is, via homotopy that is the identity on the $G_k$ coordinate and moves in a straight line segment in the $\Delta^k$ coordinate).

This argument has probably been discovered many times and is well-known, I fairly recently found out that it does appear in print at least in Segal's Classifying Spaces and Spectral Sequences, proposition 4.1. It doesn't use that $M$ is a manifold, just that the open cover has a continuous subordinate partition of unity, so for example it works for any cover of a paracompact space $M$. Even without subordinate partitions of unity, for any open cover $U_i$ of any any space $M$ the homotopy groups of $M$ and $B \mathcal{G}$ are the same: these two spaces might not be homotopy equivalent in this general setting, but they are still weakly homotopy equivalent. See Dugger and Isaksen’s Hypercovers in Topology, theorem 2.1. This simplicial space $\mathcal{G}$ is called a Čech cover, and Dugger and Isaksen's paper also deals with the more general notion of a hypercover.