Definition: Consider all Riemannian metrics on a topological surface $S$, which are classified by the conformal equivalence relation, {Riemmanian metrics on $S$}/~ , where each equivalence class is called a conformal structure .
Defintion: Suppose $g_1$, $g_2$ are two metrics on a manifold $M$, if $g_1$= $e^{2u}(g_2)$, u:$M \to \mathbb{R} $, then $g_1$ and $g_2$ are conformal equivalent.
A conformal structure based on conformal metrics would be "made up" of such metrics.
I'll register a shorter coordinate-free proof. Let the Koszul formula be written as $$2g(\nabla_XY,Z) = A(X,Y,Z) + B(X,Y,Z),$$where $A$ is the part of the Koszul formula containing directional derivatives and $B$ is the part containing Lie brackets. We know that any two connections differ by a tensor, so write $\widetilde{\nabla}_XY = \nabla_XY + T_XY$ -- the goal is to find $T$, and we know that $$2\widetilde{g}(\widetilde{\nabla}_XY,Z) = \widetilde{A}(X,Y,Z) + \widetilde{B}(X,Y,Z).$$Clearly $\widetilde{B}(X,Y,Z) = {\rm e}^{2f}B(X,Y,Z)$, while $$ X\widetilde{g}(Y,Z) = X({\rm e}^{2f})g(Y,Z) + {\rm e}^{2f}X(g(Y,Z))$$says that $\widetilde{A}(X,Y,Z) = X({\rm e}^{2f})g(Y,Z) + Y({\rm e}^{2f})g(X,Z) - Z({\rm e}^{2f})g(X,Y) + {\rm e}^{2f}A(X,Y,Z)$. Thus $$2\widetilde{g}(\widetilde{\nabla}_XY,Z) = X({\rm e}^{2f})g(Y,Z) + Y({\rm e}^{2f})g(X,Z) - Z({\rm e}^{2f})g(X,Y) + 2{\rm e}^{2f}g(\nabla_XY,Z).$$Evaluating $X({\rm e}^{2f}) = 2{\rm e}^{2f}\,X(f)$, etc., and simplyfying $2{\rm e}^{2f}$ on everything, we get $$g(\nabla_XY + T_XY,Z) = X(f)g(Y,Z) + Y(f)g(X,Z) - Z(f)g(X,Y) + g(\nabla_XY,Z).$$Eliminate $\nabla_XY$ from the above and use the definition of $g$-gradient to write the right side in the form $g({\rm something}, Z)$, obtaining $$g(T_XY,Z) = g(X(f)Y + Y(f)X - g(X,Y){\rm grad}(f), Z).$$This means that $$T_XY = X(f)Y + Y(f)X - g(X,Y){\rm grad}(f)$$and hence $$\widetilde{\nabla}_XY = \nabla_XY +X(f)Y + Y(f)X - g(X,Y){\rm grad}(f).$$
Best Answer
Short answer till you ask for more details.
(1) yes it is true for arbitrary differentiable functions $u$, we just require the factor to be positive and the exponential do it.
(2) if you to get an equivelence relation defined by $$f \ \ \text{is related to}\ \ g \ \ \ \text{if and only if}\ \ \ \ f=e^{2u}g$$ It seems to be true:
and so on. I hope it helps out.