I am having issues computing Riemann tensor of 2D manifold. It should be very simple, but I cannot see the mistake I am making.
Assume a 2D manifold with metric in very simple form $$\tilde{g}=\frac{1}{f(x)^2}(dx^2+dy^2).$$
Since this is evidently conformally flat (indeed all 2D manifolds are, but here one can see it easily), I wanted to use this to my advantage. For example from https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Conformal_change (employing notation I can easily TeX from there) for conformal transform $\tilde{g}=e^{2\varphi}g$ implies change of Riemann tensor
$$R(\tilde{g})=e^{2\varphi}(R(g)-g\bar\wedge{}T)$$
where $\bar\wedge$ denotes Kulkarni–Nomizu product and
$$T=\nabla{}d\varphi - d\varphi\,d\varphi+\frac{1}{2}|d\varphi|^2g.$$
where the norm and $\nabla$ correspond to $g = dx^2 + dy^2$, which is flat. Therefore I believe that also $R(g) = 0$.
Since $\varphi=-\log{f(x)}$, I believe that $$d\varphi = -\frac{f'}{f}\,dx,$$ $$\nabla{}d\varphi = (-\frac{f''}{f}+\frac{f'^2}{f^2})\,dx\,dx,$$ and $$|d\varphi|^2 = \frac{f'^2}{f^2}.$$ Thus $$T=-\frac{f''}{f}\,dx\,dx + \frac{1}{2}f'^2\tilde{g}$$
Therefore it seems, that $$R(\tilde{g})=-\frac{1}{2}f'^2 \tilde{g}\tilde{\wedge}\tilde{g} +\frac{f''}{f}\tilde{g}\tilde\wedge\,dx\,dx.$$
But it is common wisdom, that all 2D manifolds have Riemann tensor of the form $$R(\tilde{g})=K \tilde{g}\tilde\wedge{}\tilde{g}$$ only.
Any idea, what am I doing wrong? What about the additional term? Which assumption is wrong?
One can notice that $\bar{g}\bar\wedge{}dx\,dx = g_{yy}(dx\,dy\,dx\,dy + dy\,dx\,dy\,dx)$ and $\bar{g}\bar\wedge{}dy\,dy = g_{yy}(dy\,dx\,dy\,dx+dx\,dy\,dx\,dy)$, while also $g_{xx} = g_{yy}$. Therefore, thanks to linearity of $\bar\wedge$ in its second argument, it holds $$2\bar{g}\bar\wedge{}dx\,dx = \bar{g}\bar\wedge(dx\,dx + dy\,dy) =f^2 \bar{g}\bar\wedge\bar{g}.$$
It is a feature of low dimension and diagonality of $\bar{g}$, which together with symmetries of $\bar\wedge$ largely limits number of components of $\bar{g}\bar\wedge{}A$ for any symmetric 2-form $A$.
Therefore $$R(\bar{g}) = \frac{1}{2}(-f'^2+f\,f'')\bar{g}\bar\wedge{}\bar{g},$$ as I wanted to prove. Note that one can easily deduce solutions for a unit sphere fow which it must hold $-f'^2+ff'' = 1$ (to keep scalar curvature positive this implies $f(x) = \cosh(x)$ and indeed correponds to $\tanh(x)=\cos(R)$; $y=\varphi$ substitution in the usual sphere metric $g_{S^2} = dR^2 + \sin^2(R)d\varphi^2$) and for flat plane where indeed $f(x)=0$.
P.S.: this can indeed be generalized to $f$ depending also on $y$. If I write it up as a separate question I'll link it here.