The perturbed Hamiltonian is given as.
$$H=\begin{cases} H^{(0)}&\text{for }t\leq 0 \\ H^{(0)}+V(x)&\text{for }t>0\end{cases}.$$
Here $V(x)$ does not depend explicitly on time but it can depend on coordinates
If the initial state of system at $t<0$ is $|i\rangle$, then the probability of the transition from $|i\rangle$ to $|f\rangle$ for some $t_0>0$ is (calculated through time dependent perturbabtion theory)
$$P_{i\to f}(t_0)=\frac{4|V_{fi}|^2}{|E_f-E_i|^2}\sin^2\left(\frac{(E_f-E_i)t}{2\hbar}\right)\tag{1}. $$
If $V^{(0)}$ is constant, then the state does not change: only the energy eigenvalues get shifted.
So, for $t>0$,
$$(H^{(0)}+V^{(0)})\psi(x)=E\psi(x)
\\H^0\psi(x)+V^{(0)}(x)\psi(x)=E\psi(x)
\\\implies H^0\psi(x)=\epsilon\psi(x)$$ where $\epsilon=E-V^{(0)}$.
The eigenstates of the perturbed Hamiltonian does not change - only the energy gets scaled. Thus in this case, the transition probability from $|i\rangle\to|f\rangle$ for $i\neq f$ is $0$ if the system is in $|i\rangle$ for $t\leq 0$.
But suppose now $V=V(x)$ has position dependence, such that the full Hamiltonian is $H(x)=H^{(0)}+V(x)$. For $t>0$, we can use time independent perturbation theory.
If all the states are non-degenerate, then the first-order correction to $|i\rangle$ is $$|i^{(1)}\rangle=\sum_{n\neq i}\frac{V_{fi}}{E_i-E_n}|n\rangle.$$
We know that $|i^{(0)}\rangle$ (zeroth order term i.e., $|i\rangle$) is orthogonal to $|i^{(1)}\rangle$. So, the transition probability from $|i\rangle$ to $|f\rangle$ for $i\neq f$ is
$$P_{i\to f}(t_o)=\Big(\frac{|V_{fi}|^2}{(E_f-E_i)^2}\Big)\tag{2}$$
We can see that $(1)$ and $(2)$ are not same - there is no sine term in $(2)$. The transition probabilities in both $(1)$ and $(2)$ are obtained using the first order correction in the unperturbed wavefunctions but we are getting different results.
My question is: why are we getting different results using time-independent and time-dependent perturbation theory here?
Addendum
I am improving my question as I think I was not able to properly frame it.
For some $t_0>0$, the Hamiltonian is perturbed. Now at that instant we can look the problem through the perspective of time independent perturbation theory because the perturbed Hamiltonian does not depend explicitly on time.
Doing so, I get $(2)$ as my result and it the transition probability is independent of $t$. Then by making $t$ sufficiently small (greater than 0) my solution through time independent perturbation theory won't change.
Now we can see that before $t<0$ we have state $|i\rangle$ and after $t>0$ we have a perturbed wave function (upto first order) which is the superposition of basis states of unperturbed wave function $(\psi^{(1)}=\sum_nc_n|n\rangle)$.
In expression $(2)$, time dependence in the perturbed wave function can come only in its phase, not in observables or the transition probability. But from time dependent perturbation theory we are getting time dependence in the transition probability itself.
I am just expecting a jump in the coefficient of the wave function at $t=0$. Why are we getting different contradictory results?