You have a wrong understanding of quantum entanglement.
What is entanglement?
Quantum entanglement emerges naturally from the "only obvious way to do things" at the wavefunction level (distribute a wavefunction over all possibilities of two subsystems), and describes the fact that the general state of these systems cannot be "decoupled" into a pair of states for each of the subsystems.
Let me give you one example, qubits (quantum bits). These are variables which are allowed to be in the states $|0\rangle,\,|1\rangle,$ and any quantum superposition of them $\alpha|0\rangle + \beta|1\rangle$ where $\alpha,\beta\in\mathbb C$ and $|\alpha|^2 + |\beta|^2 = 1.$ All of those superpositions are "non-classical": there are experiments you can do on them which display strange outcomes.
The most famous strange outcome is to quantum-coherently map them to two different displays on a detector, the so-called "double slit experiment." If $|0\rangle$ becomes $|f_0(x)|^2$ and $|1\rangle$ becomes $|f_1(x)|^2$ then the state $\sqrt{\frac 12}|0\rangle + \sqrt{\frac 12}|1\rangle$ typically becomes $\frac 12 |f_0(x) + f_1(x)|^2,$ showing an "interference pattern". So if $|f_0(x)|^2$ is a bell curve and $|f_1(x)|^2$ is another bell curve, and $f_0$ and $f_1$ have complex phases which are not exactly the same, and the bell curves overlap, then in the overlap region you see some sort of "waviness" (intensity inhomogeneity) which is distinct from the "classical" overlapping-bell-curves result $\frac 12 |f_0(x)|^2 + \frac 12 |f_1(x)|^2.$
In the algebra of quantum mechanics, this comes from an "expectation value", which comes from flipping these symbols into their "duals"; I'm going to take that as a little too in-depth for this comment.
Now if you have two qubits, the general state becomes $\kappa |00\rangle + \lambda |01\rangle + \mu |10\rangle + \nu |11\rangle$ which is not necessarily representable in terms of a product of two individual states, $$(\alpha |0\rangle + \beta |1\rangle)\otimes(\gamma |0\rangle + \delta |1\rangle) = \alpha\gamma |00\rangle + \alpha\delta |01\rangle + \beta\gamma |10\rangle + \beta\delta |11\rangle.$$In particular, this is only possible when $\kappa\nu = \lambda\mu$ (they are both $\alpha\beta\gamma\delta$) but there is no reason to require that in the above "general" quantum state. So, we don't. That's where entanglement comes from; you have states which do not have this property.
Entanglement destroys coherence
Obviously, the product-states have a "quantum coherence" to both qubits: doing our double-slit experiment means that we see an interference pattern. Shockingly, entanglement weakens and sometimes eliminates this interference pattern. For example, the state $\sqrt{\frac 12} |00\rangle + \sqrt{\frac 12}|11\rangle$ describes an entangled state. If you pass the first qubit of this through the double-slit experiment, normal rules of quantum mechanics give the distribution $\frac 12 |f_0(x)|^2 + \frac 12 |f_1(x)|^2:$ classically overlapping bell curves!
This state also works a little like a classical hidden variable: two bags where I either put a toy in both of them or none of them. When you open up your bag, you do not know what you're going to get, but if there's a toy you know that the other person has a toy, or if there's no toy then the other person has no toy.
Entanglement can allow instantaneous quantum effects "over there".
Now suppose we split up the two qubits in this entangled $|00\rangle + |11\rangle$ state, where we've established that Alice is going to measure two overlapping bell curves with their double-slit experiment.
Suppose Bob likes wavy interference patterns. The rules of quantum mechanics allow Bob to do, on his qubit, any unitary transformation like $$\begin{align}
|0\rangle\mapsto& \sqrt{\frac 12} |0\rangle + \sqrt{\frac 12} |1\rangle\\
|1\rangle\mapsto& \sqrt{\frac 12} |0\rangle - \sqrt{\frac 12} |1\rangle.
\end{align}$$
This takes our state to:$$\sqrt{\frac 14} |00\rangle + \sqrt{\frac 14} |01\rangle + \sqrt{\frac 14} |10\rangle - \sqrt{\frac 14} |11\rangle $$
Now supposing that Bob measures his qubit as 0 or 1, then Alice must measure either the wavy interference patterns $\frac 12 |f_0(x) + f_1(x)|^2$ or $\frac 12 |f_0(x) - f_1(x)|^2.$ Bob can thereby instantaneously change, from a quantum perspective, what the outcomes of Alice's measurement are going to be.
Alice's wavefunction must change instantaneously and might even change retroactively: she may have already measured her qubit before Bob does this unitary transformation and measurement: nevertheless, to satisfy the predictions of quantum mechanics, her measurements must be consistent with Bob's manipulations.
But that can't send messages.
This thing that Bob has done is not directly visible to Alice, however. That's for a couple reasons, the first being that this only generates one photon of results on the double-slit screen, which isn't enough to see the pattern! But suppose we measure lots and lots of these qubits to try and see the pattern: then the problem is that Alice doesn't know which ones Bob measured as 0 or which ones Bob measured as 1. Since there was a 50/50 chance of Bob getting either, what Alice sees is therefore:$$\frac 14 |f_0(x) + f_1(x)|^2 + \frac 14 |f_0(x) - f_1(x)|^2 = \frac 12 |f_0(x)|^2 + \frac 12 |f_1(x)|^2.$$Alice therefore still measures two overlapping bell curves, overall!
Where are the interference patterns?! That is very simple: when Bob and Alice compare their measurements in the first case, Bob's 0-measurement can be used to "filter" Alice's patterns into $\frac 12 |f_0(x)|^2,$ the bell curve of photons which passed through only the first slit, and his 1-measurement filters the results to give $\frac 12 |f_1(x)|^2,$ the photons which passed through only the second slit.
Bob's transformation then changes how he can filter Alice's patterns: Alice's overlapping bell curves are now made up of the ones he measured $0$ for, which describe one wavy pattern, and the ones he measured $1$ for, which describe the other wavy pattern, and they add up into the non-wavy pattern.
And that's a general property of entanglement
What I've described is only one particular experiment, but its sneaky way of avoiding information-transfer is actually a very deep property of entanglement.
Entanglement manifests in "spooky" (non-classical) correlations of measurements, not in the measurements themselves.
You cannot observe the correlations without bringing the measurements back together, which is why you cannot transfer classical information via entanglement. You don't know anything about entanglement until you compare the two systems to see how they correlated.
Quantum teleportation, for example, uses an entangled state to "instantaneously" send an arbitrary quantum state from point A to point B. At least, that's the story. In fact, quantum teleportation sends it in a "garbled" way, and A must send a couple classical bits to B so that they can "ungarble" it and recover the full quantum state: these classical bits are absolutely necessary because the entangled state is fundamentally a correlation. It is still impressive that the arbitrary quantum bit, with its two-continuous-parameters of data, can be sent "mostly" through a quantum entanglement, reducing the actual data from "two infinities of bits" to just two bits, but it's not as magical as "teleportation" sounds to the layperson.
So that's the fundamental problem. Entanglement first destroys the quantum coherence (giving you something classical-looking) and then all of the things that you want to do with it happen within that classical veneer, and their quantum nature is only apparent when you compare two systems and say, "holy crap, you can't do that classically."
