Well, the linearity is really easy:
$$
(\nabla_a\nabla_b-\nabla_b\nabla_a)(f\,V^c) = f (\nabla_a\nabla_b-\nabla_b\nabla_a)V^c
$$
because the connection is assumed to be torsion free:
$$
(\nabla_a\nabla_b-\nabla_b\nabla_a)f=0
$$
(otherwise your "Ricci identity" won't work).
A slick way of doing this is to observe that the operator $(\nabla_a\nabla_b-\nabla_b\nabla_a)$ satisfies the product rule.
With regards to the minus sign, the problem is that the order of indices in the curvature tensor is important. It is better to write
$$
(\nabla_a\nabla_b-\nabla_b\nabla_a)V^c = R_{a b}{}^c{}_d V^d
$$
and then
$$
\begin{align}
(\nabla_a\nabla_b-\nabla_b\nabla_a)V_c
& = (\nabla_a\nabla_b-\nabla_b\nabla_a)(g_{cd}V^d) \\
&= g_{cd}(\nabla_a\nabla_b-\nabla_b\nabla_a)V^d\\
& =g_{cd} R_{a b}{}^d{}_e V^e \\
&= R_{a b d e} V^e = - R_{a b e d} V^e = - R_{a b}{}^c{}_d V_c
\end{align}
$$
(I have learned all this from R. Wald's "General relativity")
As far as I can see, the OP uses the difference tensor
$$
C_{a b}{}^c := - \delta_a{}^c \Upsilon_b - \delta_b{}^c \Upsilon_a + \Upsilon^c g_{a b}
$$
that appears in the expression
$$
\widehat{\nabla}_a \omega_b = \nabla_a \omega_b + C_{a b}{}^c \omega_c
$$
where $\nabla_a$ and $\widehat{\nabla}_a$ are the Levi-Civita connections of the metrics $g_{a b}$ and ${\widehat{g}}_{a b} = \Omega^2 g_{a b}$ respectively. Here $\Upsilon_a := \nabla_a \log \Omega$, that is the same thing as $\mathrm{d} \omega$ in the settings of this answer. The difference formula in the above display is just a restatement of the equation (1) in that answer in the abstract index notation.
The formula for the curvature $\widehat{R}$ of $\widehat{g}_{a b}$ should then look as
$$
{\widehat{R}}_{a b c}{}^d = R_{a b c}{}^d + 2 \, \nabla_{[a} C_{b] c}{}^d + 2 \, C_{[a| c}{}^e C_{e| b]}{}^d
$$
which is a direct application of the general difference formula:
Theorem (The difference formula for curvatures).
Let $\nabla$ and $\nabla'$ be two arbitrary connections in the tangent bundle of manifold $M$.
Let $K$ and $K'$ be the curvatures of the connections $\nabla$ and $\nabla'$ respectively, and $A$ be the difference tensor of the pair $(\nabla',\nabla)$, that is $ \nabla' = \nabla + A$. Then
$$
K' = K + \nabla \wedge A + A \wedge A
$$
Proof. For the sake of simplicity I will do the calculation for the case of torsion-free connections $\nabla$ and $\nabla'$ that is sufficient for our purposes (we deal with the Levi-Civita connections in the main question). In this situation the curvature $K$ of the connection $\nabla$ satisfies the identity $K_{a b c}{}^d \omega_d = 2\,\nabla_{[a} \nabla_{b]} \omega_c$, as we discussed here.
Writing everything out using the definitions, we get
$$
\begin{align}
K'_{a b c}{}^d \omega_d & = \nabla'_a \nabla'_b \omega_c - (a \leftrightarrow b) \\
& = \nabla'_a (\nabla_b \omega_c + A_{b c}{}^d \omega_d ) - (a \leftrightarrow b) \\
& = \nabla'_a \nabla_b \omega_c + (\nabla'_a A_{b c}{}^d) \omega_d + A_{b c}{}^d \nabla'_a \omega_d - (a \leftrightarrow b) \\
& = \nabla_a \nabla_b \omega_c + A_{a b}{}^d \nabla_d \omega_c + A_{a c}{}^d \nabla_b \omega_d \\
& \quad + (\nabla_a A_{b c}{}^d + A_{a b}{}^e A_{e c}{}^d + A_{a c}{}^e A_{b e}{}^d - A_{a e}{}^d A_{b c}{}^e) \omega_d \\
& \quad + A_{b c}{}^d ( \nabla_a \omega_d + A_{a d}{}^e \omega_e ) - (a \leftrightarrow b)
\end{align}
$$
Collecting the terms, we rewrite the above equation as
$$
\begin{align}
K'_{a b c}{}^d \omega_d & = \nabla_a \nabla_b \omega_c + \underbrace{A_{a b}{}^d \nabla_d \omega_c}_{\text{sym. in a, b}} + \underbrace{A_{a c}{}^d \nabla_b \omega_d + A_{b c}{}^d \nabla_a \omega_d}_{\text{sym. in a, b}} + \underbrace{A_{b c}{}^d A_{a d}{}^e \omega_e}_{\text{cancel}} \\
& \quad + (\nabla_a A_{b c}{}^d) \omega_d + \underbrace{A_{a b}{}^e A_{e c}{}^d \omega_d}_{\text{sym. in a, b}} + A_{a c}{}^e A_{b e}{}^d \omega_d - \underbrace{A_{a e}{}^d A_{b c}{}^e \omega_d}_{\text{cancel}} - (a \leftrightarrow b)
\end{align}
$$
The underbraced terms vanish for the reasons indicated in the subscripts, so we obtain the equality
$$
K'_{a b c}{}^d \omega_d = 2 \, \nabla_{[a} \nabla_{b]} \omega_c + 2 \, (\nabla_{[a} A_{b] c}{}^d) \omega_d + A_{[a| c}{}^e A_{|b] e}{}^d \omega_d
$$
which is the explicit form of the claim. QED.
A detailed account of this topic can be found in J.Slovak's dissertation here.
Best Answer
It is the ambiguity of the conventional abstract index notation for the covariant derivative that causes troubles here. Getting used to it over time helps :-)
We say that a tensor $T_{abc\dots}$ has the Bianchi symmetry if $T_{[abc]\dots} = 0$.
Your goal is to show that the tensor $\nabla_e R_{abc}{}^d := (\nabla R)_{eabc}{}^d$ has the Bianchi symmetry in the indices $e,a,b$, and this fact is called the Bianchi identity.
The fact $$ 2 \nabla_{[a} \nabla_{b]} \omega_c = (\nabla_a \nabla_b -\nabla_b \nabla_a)\omega_c=R_{abc}^{\;\;\;d}\omega_d $$ can be seen as the definition of the curvature $R$.
Taking the covariant derivative of $R_{abc}{}^d \omega_d$ by an application of the Leibniz rule we see that $$ \nabla_e (R_{abc}{}^d \omega_d) = (\nabla_e R_{abc}{}^d) \omega_d + R_{abc}{}^d \nabla_e \omega_d $$ which means that for any $\omega_d$ the tensor $\nabla R$ satisfies the identity $$ (\nabla_e R_{abc}{}^d) \omega_d = \nabla_e (R_{abc}{}^d \omega_d) - R_{abc}{}^d \nabla_e \omega_d $$
Now we can use the definition of $R$ and rewrite the last display as $$ (\nabla_e R_{abc}{}^d) \omega_d = 2 \nabla_e (\nabla_{[a} \nabla_{b]} \omega_c) - 2 \nabla_{[a} \nabla_{b]} \nabla_e \omega_c + R_{a b e}{}^d \nabla_d \omega_c $$ where we have used the Ricci identity $$ 2 \nabla_{[a} \nabla_{b]} t_{ec} = R_{a b e}{}^d t_{d c} + R_{a b c}{}^d t_{e d} $$
I will wait now for you to figure out that we are almost done.