$d\phi (X(f \circ \phi)Y)=X'(f) d\phi(Y)=X'(f)Y'$
as desired.
where $X(f\circ \phi) =d(f\circ \phi)(X) =(df \circ d\phi)(X) =df(d\phi(X)) =df(X')=X'(f)$
You have to notice that $(df \circ d\phi)(X)$ is a function on M , while $df(X')$ is (often) seen as a function on M'
The different views come from $df(d\phi(X))$
If $\nabla$ is any connection and $f$ a function, its Hessian with respect to $\nabla$ is $\mathrm{Hess}^{\nabla}f = \nabla \mathrm{d}f$, and one can see, after a messy calculation, that:
$$
\mathrm{Hess}^{\nabla}f(X,Y) - \mathrm{Hess}^{\nabla}f(Y,X) = \pm\mathrm{d}f\left([X,Y] - (\nabla_XY - \nabla_YX) \right)
$$
(where the $\pm$ sign is here because I don't remember the exact sign, but the computations are not that hard, just messy.) Hence, Hessians are symmetric if and only if the connection is torsion-free. This is the main motivation to consider torsion-free connections: in the euclidean space, Hessians are symmetric!
Moreover, the fundamental theorem of Riemannian geometry tells us that on a Riemannian manifold, there is a unique connexion that is torsion-free and lets the metric invariant, that is:
$$
\forall X,Y,Z, \left(\nabla_Zg\right)(X,Y) = Z\cdot g\left(X,Y \right) - g\left(\nabla_ZX,Y\right) - g\left(X,\nabla_ZY\right) = 0.
$$
(compare with the euclidean case, where $\langle X,Y\rangle ' = \langle X',Y\rangle + \langle X, Y' \rangle$.)
This theorem thus says that given any Riemannian metric $g$, there is a connection that is better than others: Hessians are symmetric and the metric is invariant under the action. We call it the Levi-Civita connexion.
If a connection is chosen, a geodesic is a parametrized curve satisfying the equation of geodesics : $\nabla_{\gamma'}\gamma' = 0$. Thus a curve $\gamma$ is a geodesic with respect to the connection, and can be a geodesic for some connection $\nabla^1$ but not for another connecion $\nabla^2$. Therefore, your question does not really have sense: we do not say that a connexion gives the least energy of a geodesic. I think you got confused, believing that being a geodesic is an intrinsic notion, but it really depends on the connection you consider.
Now, suppose $(M,g)$ is a Riemannian manifold endowed with its Levi-Civita connexion. Then if $\gamma : [a,b] \to M$ is a curve, we define its energy to be:
$$
E(\gamma) = \frac{1}{2}\int_a^b \|\gamma'\|^2
$$
and one can show that, in the space of all curves $\{\gamma : [a,b] \to M\}$ with same end points, a curve $\gamma$ is a point where the energy functional is extremal if and only if $\nabla_{\gamma'}\gamma'=0$, that is if and only if $\gamma$ is a solution of the equation of geodesics. Hence, a minimizer of the energy functional is a geodesic.
Best Answer
If $f:(S,g)\to (S',g')$ is an isometry, then define $\nabla_{X'}Y':=df\ \nabla_XY$
Show that this is LC-connection :
(1) Compatibility condition : First show that $$ X'(Y',Z')=X(Y,Z)$$
Proof : If $\frac{d}{dt}p(t)=X,\ p(0)=p$ then $$ df_p X(df_p Y, df_p Z) =\frac{d}{dt} (df Y, df Z)_{f(p(t))} = \frac{d}{dt} (Y,Z)_{p(t)} $$ since $f$ is an isometry And $\frac{d}{dt} (Y,Z)_{p(t)}= X(Y,Z)$
So $$ (\nabla_{X'}Y',Z')+(Y',\nabla_{X'}Z')=f^\ast g'( \nabla_XY,Z) + f^\ast g' (Y,\nabla_XZ) = X(Y,Z) =X'(Y',Z') $$
(2) Symmetry condition : $$ \nabla_{X'}Y' -\nabla_{Y'}X'=df (\nabla_XY-\nabla_YX)=df[X,Y]=[X',Y']$$