This is an interesting question. Jack gives a hint to the solution, but let me fill in the details.
What we want to show is that how $R(X,Y,X,Y)$ is related to the second fundamental form.The idea is regard the surface $M$ as a submanifold of $\mathbb R^3$, note that here the metric $g$ of $M$ is induced by the canonical metric of $\mathbb R^3$.
Now, the Levi-Civita connection $\nabla$ on $(M,g)$ satisfies $$(\nabla_XY)(p)=(\nabla_X^EY)(p)^{\top}$$
where $(\nabla^E)$ is the Levi-Civita connection on $\mathbb R^3$, and $v^{\top}$ means projection onto the tangent space of $M$. Let's introduce Gauss's equation in the following theorem:
Gauss Theorem: The curvature tensor $R$ of a submanifold $M\subset \mathbb R^n$ is given by the Gauss equation $$\langle R(X,Y)W,Z\rangle=\alpha(X,Z)\cdot \alpha(Y,W)-\alpha(X,W)\cdot \alpha(Y,Z)$$
Where $\alpha(X,Y)=(\nabla_X^EY)(p)-(\nabla_X^EY)(p)^{\top}=(\nabla_X^EY)(p)^{\bot}$, i.e., projecting vector onto normal space, and $(\nabla^E)$ is the canonical Levi-Civita connection on $\mathbb R^n$. One can see [this reference] 1 for details in section 1.8~1.10.
Specifically, in our case, $M$ is a two dimensional surface in $\mathbb R^3,$ where we have the parameterization $r=r(u,v)$, with spanning vector fields of tangent bundle $\{r_u, r_v\}$. Also in $\mathbb R^3$, the normal space of $M$ at a point is spanned by a unit normal vector $n$, and we can write $\alpha(X,Y)=\nabla_X^EY\cdot n$. And by Gauss theorem, we have $$-\langle R(r_u,r_v)r_u,r_v\rangle=\alpha(r_u,r_u)\cdot \alpha(r_v,r_v)-\alpha(r_u,r_v)\cdot \alpha(r_v,r_u)$$
Now the second fundamental form is given by $L=r_{uu}\cdot n, M= r_{uv}\cdot n$, and $N=r_{vv}\cdot n$ (note here I use different notation of second fundamental form as yours).
$$\alpha(r_u,r_u)=\nabla_{r_u}^Er_u\cdot n=\frac{\partial}{\partial u}r_u\cdot n=r_{uu}\cdot n=L,$$
and the others expressions are similar, thus we have
$$-\langle R(r_u,r_v)r_u,r_v\rangle=LM-N^2$$
which gives us the second fundamental form as we desired.
Here's a sketch of how the computation should go. By pulling back the metric from $M'$ to $M$, we may just consider one manifold $M$ with two Riemannian metrics, one a positive scalar multiple $a$ of the other. Let's write $\lambda = \sqrt a$.
You get orthonormal coframes for the two metrics, $\theta^1,\theta^2$ and $\tilde\theta^1,\tilde\theta^2$, related by $\tilde\theta^i = \lambda\theta^i$, $i=1,2$. Now differentiate and work out the structure equations to determine that (perhaps depending on your sign convention) the connection form
$$\tilde\omega_2^1 = \omega_2^1 + \left(-\frac{\lambda_2}\lambda\theta^1 + \frac{\lambda_1}\lambda\theta^2\right),$$
where $d\lambda = \lambda_1\theta^1+\lambda_2\theta^2$. This should lead to
$$\tilde\Omega_2^1 = \Omega_2^1 + \left(\Big(\frac{\lambda_1}\lambda\Big)_1 + \Big(\frac{\lambda_2}\lambda\Big)_2\right) \theta^1\wedge\theta^2.$$
(The multiplying factor is often written as $\lambda = e^\rho$, and then you are ending up, of course, with the Laplacian of $\rho$ as the additive factor in the curvature.)
EDIT: Time has passed. It seems obvious to me now that I dropped a term. There should be the additional term $\pm d\log\lambda\wedge\omega_2^1$ in the expression for $\tilde\Omega_2^1$.
Best Answer
We can define a new quantity: $$R(E_1,E_2,E_2,E_1):=\langle R(E_1,E_2)E_2,E_1\rangle$$ The Gauss equation allows us to say (since $E_1,E_2$ are orthonormal): $$R(E_1,E_2,E_2,E_1)=\det(S)=K$$ Here, $S$ is the shape operator, and the Gauss equation is $$R(E_1,E_2)E_2=\langle S(E_2),E_2\rangle S(E_1)-\langle S(E_1),E_2\rangle S(E_2)$$