On account of symmetry, 2d σ-models, even though beset with an infinity of counterterms, have those regiment themselves into few geometrical tensors constrained by symmetry; and thus limit themselves to tensors, such as the Ricci, and then underlie the β-function in the conventional manner: it is not zillions of couplings that evolve, it is only the geometry, and, on highly symmetric "pion" manifolds, by just a scale or two! (For the hypersphere, below, just its radius of curvature--it puffs up to noninteracting flatness, conformal invariance, with energy. cf. Tong 7.1.2)
Both are illustrated in our paper BCZ 1985, and especially in (2.41-2.49) and in Appendix A, meant to answer just these questions. But it is a long story, to which no justice can be done in this short format.
Nevertheless, your second question has a straightforward answer, implicit in Tong's notes, and, of course, section 2 of the paper cited here.
- In ε = d-2 dimensions, taking the "pion" fields φ to be dimensionless, but the bare metric to have dimension ε, to one loop rewrite your expression as
$$
(G_{\mu\nu}/\alpha')^{(0)}=M^\epsilon ( G_{\mu\nu}/\alpha' -\frac{1}{\epsilon}R^{(1)}_{\mu\nu} ).
$$
But the bare α'-full metric must be independent of the RG scale M; so operating on this equation by $M \frac{d}{dM}$ at the pole $\epsilon \to 0$ nets
$$
0= M \frac{d}{dM} \frac{G_{\mu\nu}}{\alpha'} - R^{(1)}_{\mu\nu},
$$
where the superscript (1) indicates the residue at the pole, and
$$
M \frac{d}{dM} \frac{G_{\mu\nu}}{\alpha'} = R^{(1)}_{\mu\nu}.
$$
Thus, in our scaled conventions, in a hypersphere ($R_{\mu\nu}=2 G_{\mu\nu}$), whose inverse
radius-squared, α', decreases with scale M , asymptotic freedom manifests itself: $d\alpha'/d\ln M =-2\alpha' ^2$. Asymptotically, the sphere flattens to a conformally invariant plane.