Motivation for transition probabilities in quantum mechanics In the area of the time-dependent perturbation theory, if we have $$\hat{H}(t) = \hat{H}_0 + \hat{V}(t).$$ The Schrodinger equation is $$i \hbar \frac{d | \Psi(t) \rangle}{dt} = (\hat{H}_0 + \hat{V}(t))| \Psi(t) \rangle$$ where in the Shrodinger picture we have $$|\Psi(t) \rangle = \sum_n c_n(t)e^{\frac{-iE_n}{\hbar}}| \psi_n \rangle$$ In the interaction picture we have $$i \hbar \frac{d | \Psi(t) \rangle_{I}}{dt} = \hat{V}_{I}(t) | \Psi(t) \rangle_{I}$$ where $$| \Psi(t) \rangle_{I} = \hat{U}(t,t_i)| \Psi(t_i) \rangle_I$$ we therefore get $$i \hbar \frac{d \hat{U}(t, t_i)}{dt} = \hat{V}_i(t)\hat{U}_{I}(t,t_i).$$ 
Question: Short question, why does it follow that the transition probability corresponding to a transition from an initial unperturbed state $| \psi_i \rangle$ to another unperturbed state $| \psi_f \rangle$ is $$P_{if}(t) = |\langle \psi_f | \hat{U}_{I}(t,t_i)| \psi_i \rangle |^2$$ also why is the transition probability in terms of the expansion coefficients given by $$P_{if}(t) = |c_{f}^{0}+c_{f}^{1} +...|^2?$$ Are these postulates?
Thanks.
 A: Question 1 :
First of all let us remember the link between the Schrödinger and Interaction Picture : $$ |\psi(t)\rangle_I  = e^{i\hat{H_0}t/\hbar} |\psi(t)\rangle_S \quad \text{and} \quad \hat{O}_I(t) = e^{i\hat{H_0}t/\hbar} \hat{O}_S(t) e^{-i\hat{H_0}t/\hbar}$$
If I understand well your notation $| \psi_f \rangle$ and $| \psi_i \rangle$ are given unperturbed states (that is eigenvectors of $\hat{H}_0$ with eigenvalues $E_f$ and $E_i$) where $f$ stands for final and $i$ stands for initial.
Well, in fact, you have already written the answer to your question :). 
Let me explain : 
The operator $\hat{U}_I(t,t_i)$ is the evolution operator in the interaction picture. By definition when you apply it to the initial state $|\psi(t_i)\rangle_I = | \psi_i \rangle$, it makes it evolve to the state at time $t$ in the interaction picture, that is $| \psi(t) \rangle_I$ : 
$$| \psi(t) \rangle_I = \hat{U}_I(t,t_i) | \psi(t_i) \rangle_
I = \hat{U}_I(t,t_i) | \psi_i \rangle$$
Now, the probability to measure the state $| \psi(t) \rangle_I$ as $| \psi_f \rangle$ is also by definition the scalar product squared : $$ P_{if}(t) = |\langle \psi_f | \psi(t)\rangle_I|^2 $$ 
Hence : $$ P_{if}(t) = |\langle \psi_f| \hat{U}_I(t,t_i) | \psi_i \rangle|^2 $$ 
As a remark this is equal to : 
$$ P_{if}(t) = |\langle \psi_f| \psi(t) \rangle_S|^2 = |\langle \psi_f|e^{-i(\hat{H}_0+\hat{V}_i(t))t/\hbar}| \psi_i \rangle|^2 $$
Question 2
Using the remark above :
$$ P_{if}(t) = |\langle \psi_f| \psi(t) \rangle_S|^2  $$ 
$$ P_{if}(t) = |\langle \psi_f | \sum_n c_n(t) e^{-iE_nt/\hbar}  | \psi_n \rangle|^2$$
$$ P_{if}(t) = |c_f(t) e^{-iE_ft/\hbar}|^2 = |c_f(t)|^2$$
