= Solution
Define the <interaction picture> by $|\psi_I(t)\rangle=e^{iH_0t/\hbar}|\psi_S(t)\rangle$. Differentiating and using the <Schrödinger equation> cancels the free Hamiltonian:
$$
i\hbar\frac d{dt}|\psi_I(t)\rangle
=V_I(t)|\psi_I(t)\rangle,\qquad
V_I(t)=e^{iH_0t/\hbar}V(t)e^{-iH_0t/\hbar}.
$$
Integrating once and replacing the state in the integral by its initial value gives
$$
|\psi_I(t)\rangle=|a\rangle-\frac i\hbar\int_0^tV_I(t')|a\rangle\,dt'
+O(V^2).
$$
Since $\langle b|a\rangle=0$, the first-order transition amplitude is $-i/\hbar$ times the integral of $\langle b|V(t')|a\rangle e^{i(E_b-E_a)t'/\hbar}$. Its modulus squared gives
$$
\boxed{P_{a\to b}(t)=\frac1{\hbar^2}
\left|\int_0^t\langle b|V(t')|a\rangle
e^{i(E_b-E_a)t'/\hbar}\,dt'\right|^2}
$$
to second order in the perturbation.
For the two-state example, the matrix element is $vt'$ and the free energies coincide. Thus $P_{1\to2}=v^2t^4/(4\hbar^2)$ to order $v^2$. All Hamiltonians commute at different times because they are combinations of $I$ and the same <Pauli matrix> $\sigma_1$. Therefore the ordinary exponential of their time integral solves the equation: differentiating it gives $-(i/\hbar)(EI+vt\sigma_1)$ times itself, and its value at zero is $I$. Since $\sigma_1^2=I$, it is
$$
e^{-iEt/\hbar}\left[\cos\left(\frac{vt^2}{2\hbar}\right)I
-i\sin\left(\frac{vt^2}{2\hbar}\right)\sigma_1\right].
$$
Starting from state 1, the exact result and the approximation condition are
$$
\boxed{P_{1\to2}(t)=\sin^2\left(\frac{vt^2}{2\hbar}\right),\qquad
\left|\frac{vt^2}{2\hbar}\right|\ll1.}
$$
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