How to apply QFT to a quantum state in superposition?

QFT on any Superposition (Linear Combination of Basis States) can be applied using Linearity.

$$QFT_n|psirangle = sum_{k=0}^{2^n-1}a_kQFT_n|krangle$$

Hence $QFT_4|psirangle$ where $|psirangle = frac{1}{2}(|0000rangle + |0100rangle + |1000rangle + |1100rangle)$ is $$QFT_4(frac{1}{2}(|0000rangle + |0100rangle + |1000rangle + |1100rangle)) = frac{1}{2}(QFT_4|0000rangle + QFT_4|0100rangle + QFT_4|1000rangle + QFT_4|1100rangle) = frac{1}{2}(QFT_4|0rangle + QFT_4|4rangle + QFT_4|8rangle + QFT_4|12rangle) = frac{1}{2}(frac{1}{4}sum_{k=0}^{15}omega_N^{ktimes 0}|krangle + frac{1}{4}sum_{k=0}^{15}omega_N^{ktimes 4}|krangle + frac{1}{4}sum_{k=0}^{15}omega_N^{ktimes 8}|krangle + frac{1}{4}sum_{k=0}^{15}omega_N^{ktimes 12}|krangle ) =frac{1}{8}sum_{k=0}^{15}(omega_N^{ktimes 0}+omega_N^{ktimes 4}+omega_N^{ktimes 8}+omega_N^{ktimes 12})|krangle $$

Here $omega_N = e^{frac{i2pi}{2^4}} = e^{frac{ipi}{8}}$, therefore $omega_N^0 = 1$, $omega_N^4 = e^{frac{ipi}{2}} = i$, $omega_N^8 = e^{ipi}=-1$ and $omega_N^{12} = e^{frac{i3pi}{2}}=-i$.

Thus $QFT_4|psirangle$ is

$$ QFT_n|psirangle = frac{1}{8}sum_{k=0}^{15}(omega_k^{ktimes 0}+omega_k^{ktimes 4}+omega_k^{ktimes 8}+omega_k^{ktimes 12})|krangle = frac{1}{8}sum_{k=0}^{15}(1^{k}+i^{k}+(-1)^{k}+(-i)^{k})|krangle $$

This sum $(1^{k}+i^{k}+(-1)^{k}+(-i)^{k})$ is $4$ when $k$ divides 4 otherwise its 0.

Thus $QFT_4|psirangle$ is

$$QFT_n|psirangle = frac{1}{8}sum_{k'=0}^{3}(4)|4k'rangle = frac{1}{2}(|0rangle + |4rangle + |8rangle + |12rangle) = frac{1}{2}(|0000rangle + |0100rangle + |1000rangle + |1100rangle) $$

Here I noticed that our initial state is an eigenvector of $QFT_4$ with an eigenvalue $1$. If I noticed this before I could have directly written the answer.

Nonetheless, I hope this helps.