How to get a qubit into superposition?

The Hadamard gate put the states $|0rangle$ and $|1rangle$ into uniform superposition but not for any arbitrary state.

You can check that: $H|0rangle = dfrac{1}{sqrt{2}} begin{pmatrix} 1& 1 1 & -1 end{pmatrix} begin{pmatrix} 1 0end{pmatrix} = dfrac{1}{sqrt{2}}begin{pmatrix} 1 1 end{pmatrix} = dfrac{|0rangle + |1rangle}{sqrt{2}}$ and similarly, you can show that $H|1rangle = dfrac{|0rangle - |1rangle}{sqrt{2}}$. In each case, the probability to see a $|0rangle$ or $|1rangle$ is $big| 1/sqrt{2}big|^2 = 1/2. $

If you have an arbitrary single qubit state, $|psi rangle = alpha|0rangle + beta|1rangle $ then you can show that $H|psirangle = (alpha + beta)|0rangle + (alpha - beta)|1rangle$, so the probability of seeing the state $|0rangle$ is $big| alpha + beta|^2$, and the probability of seeing the state $|1rangle$ is $big| alpha - beta big|^2$, thus they are not the same. So applying the Hadamard gate to an arbitrary state does not bring it to a uniform superposition state.

Of course, you can find a Unitary matrix $U$ that maps the state $|psirangle$ to the uniform superposition state $dfrac{|0rangle + |1rangle}{sqrt{2}}$. But from all your previous questions, it seems like you are using Qiskit for the work that you are doing. What you can do is to reset the particular qubit you are interesting in to the $|0rangle$ state through the reset option, then apply the Hadamard gate. This is probably the easiest. Below is an example of how to do reset:

from qiskit import QuantumRegister, ClassicalRegister, QuantumCircuit
qreg_q = QuantumRegister(1, 'q')
creg_c = ClassicalRegister(1, 'c')
circuit = QuantumCircuit(qreg_q, creg_c)
circuit.ry(1.213, qreg_q[0])  #A quantum state that is not in uniform superposition
circuit.reset(qreg_q[0]) # reset the qubit to |0>
circuit.h(qreg_q[0]) # Apply the Hadamard  gate 
print(circuit)

     ┌───────────┐     ┌───┐
q_0: ┤ RY(1.213) ├─|0>─┤ H ├
     └───────────┘     └───┘
c: 1/═══════════════════════