How does an ideal op amp amplify a voltage input when the voltage difference is 0?

A transfer function of closed loop system is:

\$H(s) = \dfrac{F(s)}{1+F(s)}\$

^{simulate this circuit – Schematic created using CircuitLab}

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A transfer function of closed loop system with a feedback is:

\$H(s) = \dfrac{F(s)}{1+F(s)\cdot G(s)}\$

^{simulate this circuit}

Now let's suppose you have a opamp buffer - the output is feedback to inverting input, as case 1 in this answer. F(s) = G (Gain), you can notice that output is never equal to the voltage on non-inverting input, but very close.

^{simulate this circuit}

The error is :

\$\varepsilon = \dfrac{1}{1+G}\$

So, the voltages on both inputs are never equal, but very close, and higher gain makes them closer. Since the opamp's gain is insane high, for the sake of simplicity we can say that voltage of inverting input is equal to the voltage of the non-inverting input. If you ground the non-inverting input, then you get a virtual ground on the inverting input, but it is never zero voltage for real. And as you already discovered it also can't be zero, else no signal would come out.

It's a Platonic ideal of an op-amp; it's not a thing that could ever exist in the real world.

So you say to yourself "I'll just assume that my system is stable" (real op-amp circuits aren't, always), and "I'll just assume that my output does whatever it needs to hold the input voltage difference at zero" (real op-amps don't, quite), and "I'll just assume my op-amp is infinitely fast" (this is impossible). And then you work really hard to just stop worrying about it because it's all pretend, to make the math easier, and because for a lot of real-world problems it's Good Enough^TM.

Later on, if you pursue analog circuit design, they'll start expecting you to solve problems like "how accurate is the circuit if the gain is only \$10^5\$?" and "how do the op-amp's very real bandwidth limitations affect the circuit performance?" But those are much harder questions to answer if you can't go back to that ideal op-amp touchstone and figure out what the ideal circuit would do.

Uin = Uout / A, so if A is big, Uin is small. That's it.

Just like it took ages to "invent" zero, the concept of "infinite" is also not that easy to understand. For now, just consider that an (ideal) opamp has an amplification that is "very large", so that Uin is "very small". You may consider Uin "zero" as long as you understand that it is not really zero, but small enough so that you are having difficulty measuring it, and big enough so that it gets amplified to get Uout.