In the setting as stated in the question, this is a sort of muscle exercise, and the answer will not look nice at all. The things get a little simpler if we put
$$
f=e^{2 \omega}
$$
Using the Koszul formula we obtain
$$
\nabla' _X Y = \nabla _X Y + (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{1}
$$
Any two linear connections $\nabla'$ and $\nabla$ are related to each other by
$$
\nabla' _X Y = \nabla _X Y + S(X,Y) \tag{2}
$$
where $S(X,Y)$ is a (1,2)-tensor, the difference tensor of the pair of connections. In our case, the connections are torsion free, and thus the difference tensor is symmetric:
$$
S(Y,X)=S(X,Y)
$$
Moreover, for torsion-free connections $\nabla'$ and $\nabla$ the corresponding curvature endomorphisms are related via their difference tensor $S$ as
$$
R'(X,Y)Z = R(X,Y)Z + \nabla_X S(Y,Z) - \nabla_Y S(X,Z)+S(X,S(Y,Z))-S(Y,S(X,Z)) \tag{3}
$$
where by definition
$$
\nabla_X S(Y,Z):= \nabla_X (S(Y,Z)) - S(\nabla_X Y,Z) - S(Y, \nabla_X Z) \tag{4}
$$
Now it is straightforward to substitute
$$
S(X,Y) = (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{5}
$$
into (3) and obtain the expression for the conformal transformation of the curvature endomorphisms. It can be found in W.Kühnel's "Differential Geometry", p.349, and I reproduce the formula with some modifications:
$$
\begin{align}
R'(X,Y)Z &= R(X,Y)Z + g(\nabla_X \operatorname{grad} \omega,Z)Y - g(\nabla_Y \operatorname{grad} \omega,Z)X\\
&+ g(X,Z)\nabla_Y \operatorname{grad} \omega - g(Y,Z)\nabla_X \operatorname{grad} \omega\\
&+ (Y \omega)(Z \omega)X - (X \omega)(Z \omega)Y\\
&- g(\operatorname{grad} \omega, \operatorname{grad} \omega)[g(Y,Z)X - g(X,Z)Y]\\
&+ [(X \omega)g(Y,Z)-(Y \omega)g(X,Z)]\operatorname{grad} \omega
\end{align} \tag{6}
$$
Using this one can now find expressions for the Ricci curvature, the scalar curvature etc.
Formula (6) suggests the outstanding role of $\operatorname{grad} \omega$ in conformal transformations. This actually is an appearance of $d\omega$ which is canonically identified in the Riemannian geometry with $\operatorname{grad} \omega$ by virtue of the musical isomorphisms.
To get more insights one can observe that the difference tensor $S(X,Y)$ has a more symmetric presentation in the "covariant" form:
$$
g(S(X,Y),Z) = d\omega (X) g(Y,Z) + d\omega (Y) g(X,Z) - d\omega (Z) g(X,Y) \tag{7}
$$
From this point it is already not that far from realizing that the (0,4)-curvature tensor will have a nicer expression, and in fact it is so.
An ultimate understanding of the conformal transformation of the curvature is obtained by analyzing the algebraic properties of the curvature tensor, the direction that is better covered in the language of the representation theory.
A canonical list used in the references is given in A.Besse's "Einstein manifolds" on pp. 58-59.
As for the sectional curvature, I think that it would be a very straightforward calculation using formula (6) and the definition
$$
K = \frac{g(R(X,Y)Y,X)}{g(Y,Y)X - g(X,X)Y} \tag{8}
$$
I hope that this answer clarifies the things a little.
This is somewhat delicate to do with complete detail and involves exactly the kind of issues on which Do Carmo doesn't bother to expand.
Let $\gamma : [0,1] \rightarrow M$ be a smooth curve and let $P,P'$ be two smooth parallel vector fields along $\gamma$. Define $f(t) = g(P(\gamma(t)), P'(\gamma(t)))$. The function $f : [0,1] \rightarrow M$ is smooth. We will show that the derivative $f'(t)$ is identically zero.
Let $t_0 \in [0,1]$. Choose some vector field $Z \in \Gamma(M)$ such that $Z(\gamma(t_0)) = \dot{\gamma}(t_0)$. Consider three cases:
(1) $\dot{\gamma}(t_0) \neq 0$. Then, for small enough $\epsilon > 0$, the image $\gamma([t_0 - \epsilon, t_0 + \epsilon])$ is an embedded compact submanifold of $M$ with boundary. This implies that $P,P'$ (restricted to the image) are defined on a compact submanifold of $M$ and so can be extended to vector fields $X,X' \in \Gamma(M)$, defined $\textbf{globally}$ on $M$, such that $X(\gamma(t)) = P(\gamma(t)), X'(\gamma(t)) = P'(\gamma(t))$ for $t \in [t_0 - \epsilon, t_0 + \epsilon]$. Thus,
$$ f'(t_0) = \dot{\gamma}(t_0)g(X,X') = \left. Zg(X,X')\right|_{p=\gamma(t_0)} = \left. g(\nabla_Z X, X')\right|_{p=\gamma(t_0)} + \left. g(X, \nabla_Z X')\right|_{p=\gamma(t_0)} = g(\left. \frac{DP}{dt} \right|_{t=t_0}, X'(\gamma(t_0))) + g(X(\gamma(t_0)), \left. \frac{DP'}{dt} \right|_{t=t_0}) = 0. $$
(2) $\dot{\gamma}(t_0) = 0$, but there is a sequence $t_n \rightarrow t_0$ such that $\dot{\gamma}(t_n) \neq 0$. For example, you can think about a curve $\gamma$ that traces a line segment from $t = 0$ to $t_0 = \frac{1}{2}$, slows down smoothly as $t$ approaches $\frac{1}{2}$, and then turns back on its trace. In this case, you can't necessarily extend the vector fields $P, P'$ to $\textbf{global}$ vector fields, but you get from continuity of $f'$ and from the previous analysis that $f'(t_0) = 0$.
(3) $\dot{\gamma}(t) \equiv 0$ in a neighborhood of $t_0$. This implies that $\gamma(t)$ is constant around $t_0$. Since parallel transport along a constant curve is constant, the function $f(t)$ itself is constant around $t_0$ and in particular $f'(t_0) = 0$.
Best Answer
Take a look at the last big displayed equation under "formal definition" here. It shows you Gauss's explicit form for a Levi-Civita connection in terms of the metric. Since you know how the metric transforms under an isometry, and how a Lie bracket transforms under a diffeomorphism, working out how the connection transforms under an isometry amounts to putting those ingredients together.
http://en.wikipedia.org/wiki/Levi-Civita_connection