It does seem like you have some misconceptions. You don't have to measure them simultaneously, in fact the whole idea of "being simultaneous" turns out to be subjective and observer dependent in relativity.
But the real issue is that there are many measurements you can do. For instance you could measure the z component of spin, or you could measure the y component of spin, or you could measure the x component of spin. But you can't measure all three at once, even for a single particle.
Let's go over measurement. Let's say a beam is travelling in the positive y direction and the beam has some thickness in the x and z directions and goes through a region of magnetic field where the magnetic field is in the z direction, is inhomogeneous in the z direction, and only is strong for one brief region near $y=0.$ I just described the setup of a Stern-Gerlach device. The result is that an incoming beam might be one beam coming in purely in the y direction with a superposition of two spin states (a state with a positive z component of spin and a different state with a negative z component of spin, i.e. the z component of the spin is positive or negative respectively in those two state). As the beam goes through, the Schrödinger equation (which is how things evolve in time in quantum mechanics) predicts that the beam splits into two beams, one angled up in the z direction and one that angles down in the z direction. But the spin becomes purely up in one of the beams and becomes purely down in one of the beam.
A general superposition of spin states for a spin half particle can be visualized by a unit 3d vector, so it can come in with any spin vector, but one outgoing beam has a spin vector that points straight up and the other outgoing beam has a spin vector that points straight down.
So you can imagine visualizing this process by pulling out a piece of paper and having a positive $z$ axis going to the top of the paper and the positive y axis going to the right. Then draw two parallel horizontal lines on the left side that's the $z$ width on the incoming beam. Then near $y=0$ angle the top line upwards and the bottom one downwards and add a sideways v so that the whole thing looks like a Y that fells down to the right. So to he right there are two beams each has a spin that points in the $\pm z$ direction.
That's one particle, and measuring in just the z direction. Imagine picking a vertical cross section like $y=-2$ you would see some thickness in the z direction and a spin vector pointing any direction whatsoever.
Then you could pick a vertical cross section like $y=+2$ and you would see some thickness in the z direction and a spin vector pointing like $+\hat z$ then farther down you would see some thickness in the z direction and a spin vector pointing like $-\hat z.$
For slices in between you'd see some overlap as the beam spreads before breaking into two and as the spin vectors starts to polarize along the stream lines that end up in the two branches.
So practice seeing this with the y axis just imagine a movie where over time a line segment in the z direction gets longer and then separates into two pieces that are then moving away from each other and imagine the spin polarizes itself so that when the pieces are separate they each have a spin in the $\pm \hat z$ direction.
Why did we go to so much work? Because the wavefunction is another a wave in space when you have more than one particle. $\Psi=\Psi(t,x,y,z)$ only works for one particle. For two particles you have $\Psi=\Psi(t,x_1,y_1,z_1,x_2,y_2,z_2).$
This is probably news to you, the wavefunction of quantum mechanics doesn't assign a complex number to locations in space it assigns complex numbers to configurations of all the particles. So for instance $(t,x_1,y_1,z_1,x_2,y_2,z_2)$ corresponds to the configuration of particles at time t where particle one is at $(x_1,y_1,z_1)$ and particle two is at $(x_2,y_2,z_2).$ So if $\Psi(t,0,0,0,10m,0,0)$ is nonzero, that (very roughly) corresponds to a probability of particle one at the origin and particle two being 10m over in the x direction.
So the wavefunction can describe whole configurations of all the particles.
So now imagine you want to measure two particles. Each can be a beam heading in the positive y direction each with some initial thickness in the x direction and the z direction, just at say one beam is near $x=0m$ and the other is near $x=10m$ so they are parallel to each other. We are going to ignore the x and y directions because we need to visualize two z directions.
So imagine it is initially a square in the $z_1,z_2$ plane. Let's draw that on our paper with with $z_1$ going right and $z_2$ going up.
If we measure just particle one then the square gets longer in the left right direction, becomes a rectangle then develops a vertical line somewhere down the middle and then the two new rectangles start to move left and right away from each other.
If we measure just particle two then the square gets taller in the up down direction becomes a rectangle then develops a horizontal line somewhere across the middle and then the two new rectangles start to move up and down away from each other.
Where does this line form? It depends on the original spin. If it was all spin up coning in it forms on the far edge, i.e. the whole beam just deflects without forming two beams. If the incoming spin is an equally sized superposition of spin up and spin down then the line forms right in the exact middle and equally sized beams go left and right. For spin that is more spin up than spin down the line forms so the two beams have relative sizes related to how much more spin up the incoming beams had compared to spin down. (For experts, the L2 norm of each beam adds up to the L2 norm of the incoming beam with the sizes relative to the L2 size of the projection of the original spin onto the spin eigenstates.)
So now we understand what measurement look like, and we can visual in real time what happens in a real measurement. This is great (and most people don't bother to learn this skill).
So let's look at an entangled state where the spins are entangled to have the same spin as each other. To do this, it happens to require that the beam splits into two equal beams.
So what happens if you measure particle one first? Recall that the square gets longer in the left right direction, becomes a rectangle then develops a vertical line somewhere down the middle and then the two new rectangles start to move left and right away from each other. But in this case the line forms right in the middle. And remember that the two separated beams have the spin get polarized? Well, since the spins are entangled both spins become up in one rectangle and both spins become down in the other rectangle. But the rectangles didn't get taller, to the person standing by the other beam it still looks like a regular width beam. Now if you measure the second beam, each of those rectangles just starts deflecting up or down because each of them is purely spin up or purely spin down, and that is what a spin measurement does on something that is all spin up (it deflects it up) or all spin down (it deflects it down). So you end up with two squares in the $z_1,z_2$ plane.
What if you measured the spin of particle two first? Then the initial square grows taller in the up down direction becomes a rectangle then develops a horizontal line right across the exact middle and then the two new rectangles start to move up and down away from each other. But then remember that the two separated beams have the spin get polarized? Well, since the spins are entangled both spins become up in one rectangle and both spins become down in the other rectangle. But the rectangles didn't get wider, to the person standing by the other beam it still looks like a regular width beam. Now if you measure the first beam, each of those rectangles just starts deflecting left or right because each of them is purely spin up or purely spin down, and that is what a spin measurement does on something that is all spin up (it deflects it right) or all spin down (it deflects it left). So you end up with two squares in the $z_1,z_2$ plane.
In both cases the squares are in the upper-right corner of the $z_1,z_2$ plane and the lower-left corner of the $z_1,z_2$ plane. This is because it was entangled to have the same spin. If they were entangled to have opposite spins they would have ended up in the upper-left and the lower-right.
What if you measure both beams at the same time? It evolves from one square into two squares in those opposite corners as the spin polarizes to have the spins be correlated.
Now, you talked about signalling.
Let's say you measured particle one first. Then a vertical line appeared and two squares separated in the left right direction. What does that look like to the person by the second beam? Nothing. All the changes happened to the z of particle one, to someone that can only see particle one any motion left-right is undetectable. That whole getting longer and separating and moving left and right was all about the configuration where the only thing changing was the coordinate of the particle you can't see. But what about the spin? That changed, right? But you can't see spin. In fact these separating beams is how you detect spin.
All the second person can see is that they have a beam and that later they can get two beams. They have no idea whether the beam is separated in the direction of some other particle that is far away from them. One person can only see one particle, the one with an x coordinate near them (I know we were drawing the x coordinates but that how you know which particle you are measuring, one person was near $x=0m$ the other person was near $x=10m$ and each had just one particle near them).
So one person can only see the up down direction in our $z_1,z_2$ plane and the other person can only see the left right direction in our $z_1,z_2$ plane. So they can't send information to each other.
Which is great, since for other reasons we happen to know that simultaneous is not an objective thing.