TL;DR: There are 8 possible outcomes, each with equal probability of being read. The final state has q[1] read as 0 in 4 of the outcomes, and read as 1 in the other 4 outcomes. Since you are only measuring q[1], you only see the two results, one with q[1] as 0 and one with q[1] as 1. They are both around 50% because each outcome had an equal probability of being read.
Let's go through your circuit and what the state should be at each step.
X-Gate on q[2]
The state will now be |100$\rangle$
Just a simple bit flip on q[2], from 0 to 1.
H-Gate on q[2] and H-Gate on q[1]
After these gates, the state will be $\frac{1}{\sqrt4}$|000$\rangle$ + $\frac{1}{\sqrt4}$|010$\rangle$ + $\frac{1}{\sqrt4}$|100$\rangle$ + $\frac{1}{\sqrt4}$|110$\rangle$
q[2] and q[1] both get put into superposition. They both can be either 0 or 1, leading to 4 possible outcomes.
CNOT from q[1] to q[0]
After this gate, the state will be $\frac{1}{\sqrt4}$|000$\rangle$ + $\frac{1}{\sqrt4}$|011$\rangle$ + $\frac{1}{\sqrt4}$|100$\rangle$ + $\frac{1}{\sqrt4}$|111$\rangle$
q[1] and q[0] become entangled. q[0]'s outcome depends on q[1]. The two will either be 00 or 11, and since q[2] is still in superposition, there are still 4 possible outcomes.
H-Gate on q[2] and H-Gate on q[0]
After these gates, the state will be $\frac{1}{\sqrt4}$|100$\rangle$ + $\frac{1}{\sqrt4}$|101$\rangle$ + $\frac{1}{\sqrt4}$|110$\rangle$ + $\frac{1}{\sqrt4}$|111$\rangle$
The H-Gate on q[2] cancels the first one out, leaving q[2]'s outcome as 1. The H-Gate on q[0] puts it in superposition, meaning it can be either 0 or 1. There are still 4 outcomes, because although we lost two possible outcomes with q[2] going back to just being 1, we gained two possible outcomes since q[0] can be either 0 or 1.
CNOT from q[2] to q[0]
After this gate, the state will still be $\frac{1}{\sqrt4}$|100$\rangle$ + $\frac{1}{\sqrt4}$|101$\rangle$ + $\frac{1}{\sqrt4}$|110$\rangle$ + $\frac{1}{\sqrt4}$|111$\rangle$
This gate essentially does not change the state. What ends up happening is when q[2] is 1, a bit flip will be applied to q[0]. However, at the moment q[2] is always 1, so each possible outcome will have q[0]'s bit flipped. Since each outcome has an equal probability of being read, the bit flip just provides us with the state possible outcomes, each with the same probabilities.
H-Gate on q[2] and H-Gate on q[0]
After these gates, the state will be $\frac{1}{\sqrt4}$|000$\rangle$ + $\frac{1}{\sqrt4}$|011$\rangle$ + $\frac{1}{\sqrt4}$|100$\rangle$ + $\frac{1}{\sqrt4}$|111$\rangle$
The H-Gate on q[2] puts it back in superposition, meaning it can be either 0 or 1. The H-Gate on q[0] cancels out the first one, meaning it is either 0 or 1 depending on q[1]. q[0] and q[1] will either be 00 or 11, and q[2] will either be 0 or 1, providing us with 4 possible outcomes.
H-Gate on q[1]
After this gate, the state will be $\frac{1}{\sqrt8}$|000$\rangle$ + $\frac{1}{\sqrt8}$|001$\rangle$ + $\frac{1}{\sqrt8}$|010$\rangle$ + $\frac{1}{\sqrt8}$|011$\rangle$ + $\frac{1}{\sqrt8}$|100$\rangle$ + $\frac{1}{\sqrt8}$|101$\rangle$ + $\frac{1}{\sqrt8}$|110$\rangle$ + $\frac{1}{\sqrt8}$|111$\rangle$
This final H-gate sets q[1]'s outcome as either 0 or 1, and this change won't affect q[0]'s outcome. q[0]'s outcome is still either 0 or 1 depending on what q[1] at the CNOT gate. q[2] is still in superposition, so it is either 0 or 1 as well. Since all 3 of the qubits can be either 0 or 1, there are $2^3$, or 8, possible outcomes.
The last H-Gate on q[1] does not cancel out the first H-Gate (which would end up setting q[1] to only be 0) because of the entanglement between q[0] and q[1].
Unfortunately, I do not know exactly how to fix your circuit to have it do what you want it to, as I am not very knowledgable on the algorithm itself. However, this article may provide some more information on the topic.
