Can we determine what the wavefunction (states) of a particle is before we decide which measurement to make? If we are measuring spin-up or spin-down then we write the wavefunction (I think) as
$$ \require{physics} \psi = \tfrac{1}{\sqrt{2}} |{\uparrow}\rangle+\tfrac{1}{\sqrt{2}} |{\downarrow}\rangle  $$
But if we emit a particle, only to make a measurement after a long period of time and distance, and we haven't decided what to measure yet, does the particle still have a wave function? We might measure up/down, left/right or some other quantum parameter.
Would it include all possible measurements each with 0 amplitude (since they are infinite as many). And after we decide to measure up or down you have a wavefunction coalescing into two distinct states? Kind of like a sub-collapsing before the actual collapsing that happens during the measurement.

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*What can we know apriori before the measurement, and how is that encoded into the wavefunction?


*What about two entangled particles? Is there a relationship in their wavefunctions before measurements? Is a relationship like $\int \langle \psi_1 | \psi_2 \rangle = 0 $ where the two wavefunctions are "orthogonal" to each other, or something like that?
 A: If the particle starts in the state
$$ \require{physics} \psi = \tfrac{1}{\sqrt{2}} |{\uparrow}\rangle+\tfrac{1}{\sqrt{2}} |{\downarrow}\rangle  $$
and some time elapses, how the state develops depends on the hamiltonian of that particle and any other particle it may interact with. But there is no reason that the state should be very complicated after a long time, especially if there is no strong interaction with another particle.
Answering your specific questions:

But if we emit a particle, only to make a measurement after a long period of time and distance, and we haven't decided what to measure yet, does the particle still have a wave function?

No matter what, there would be a wavefunction that describes the particle. If the particle interacts with other particles and becomes entangled, the wavefunction which describes that particle would be a shared wavefunction between all entangled particles, but nonetheless you could pull the probabilities for individual particles from it too.
Maybe you consider a joint wave function not to be "a wave function for that particle". In that case you could say it doesn't have a wave function. That is up to you as it is just a matter of semantics.

Would it include all possible measurements each with 0 amplitude (since they are infinite as many).

No, given any spin state - including the one in your example - there are always infinitely many possible measurements that you can do on it, by measuring the component of its spin along any axis. However given a specific axis, only two outcomes are possible no matter what. And each outcome may have a nonzero probability.

What can we know apriori before the measurement, and how is that encoded into the wavefunction?

If we know the Hamiltonian of the system, and we know the state in which it started, we can know the probabilities of getting any particular spin result along any axis.

What about two entangled particles? Is there a relationship in their wavefunctions before measurements? Is a relationship like $\int \langle \psi_1 | \psi_2 \rangle = 0 $ where the two wavefunctions are "orthogonal" to each other, or something like that?

They don't have separate wavefunctions. Any two entangled particles have just one wavefunction that describes both of them. And as such there is no sort of orthogonality relation. Take for example the entangled singlet state used in the bell theorem:
$$\frac{1}{\sqrt{2}} (|\uparrow \downarrow \rangle - |\downarrow \uparrow \rangle)$$
This basically reads "Either the first particle has spin up along the z-axis and the second spin down, or vice versa". The minus sign encodes information about probabilities for measuring along other axes.
A: Note that your spin 1/2 wave function:
$$ \require{physics} \psi = \tfrac{1}{\sqrt{2}} |{\uparrow}\rangle+\tfrac{1}{\sqrt{2}} |{\downarrow}\rangle = |\rightarrow\rangle $$
where:
$$|\rightarrow\rangle \equiv |S=\tfrac 1 2, S_x=\tfrac 1 2 \rangle $$
That is, it is an eigenstate of the $\hat S_x$ operator.
A: It sounds like you're conceptualizing a wavefunction as a big statistical ensemble of all possible outcomes of measurements. That's not what it is. In your example, let $u$ be the spin-up wavefunction and $v$ the spin-down wavefunction. Let's not worry about the normalization factors of $1/\sqrt{2}$ for now. Then your $u+v$ is not the same wavefunction as $u-v$, nor is it the same state. The phases matter.
If the phases didn't matter, then you'd be talking about a statistical ensemble of many electrons, which is described by a density matrix, not a single wavefunction (not even a pure-state wavefunction of many electrons).
Consider the double-slit experiment, and now redefine $u$ as a wavefunction in which the wave is only emerging through the left slit, and $v$ as a wavefunction in which the wave is emerging through the right slit. Let's say $u$ and $v$ are in phase at the slits. Then $u+v$ gives an interference pattern in which there is a maximum at the center, while $u-v$ gives a minimum at the center.
Measurement doesn't have any fundamental logical significance in quantum mechanics. Wavefunctions are what they are, regardless of whether there is any measurement. Measurement is just a certain type of physical interaction in which decoherence takes place.
