Electronic – How to a diode parallel to the source can’t take the source value but take about 0.7V

The mistake depends by what you are trying to achieve.

In this case, you have the voltage source that is suppose to simulate the forward-biased diode. This means that, since the source is connected with the plus to ground, you are generating -0.7V at the non-inverting input of the OPAMP. So, a current is flowing from ground across the source, and that current depends on the output voltage and the value of the top resistance (perhaps 1 Ohm).

Then, let's look at the inverting input and the voltage divider. Since between the two OPAMP inputs there is a virtual short circuit, the central point of the divider will be -0.7V. Using two equal (1 Ohm I guess) resistors, you are causing the output to be at -1.4V. Again, the current will flow out of the ground.

Now, back again to our generator. We said that the non-inverting input is at -0.7, and the output is at -1.4V. Hence we will have a 0.7V drop over the resistor, and since (as guessed before) it's a 1 Ohm resistor, the current across the resistor and the generator/diode (since the OPAMP inputs are ideally open circuits) will be 0.7A

Conclusion

If you are trying to simulate the 0.7V drop of the forward-biased diode, it's what the supply is doing. If you are expecting to see positive voltages, it's not because of negative resistors, but because the supply has to be flipped.

Update

There are two cases, depending on the initial state:

1. The output of the OPAMP is HIGH: then, the diode is reverse biased, no current is flowing in the upper branch, and the non-inverting input is at a higher voltage than the inverting, that is always at half the output voltage. Hence, the OPAMP goes into positive saturation;

2. The output is LOW: then, the diode is forward biased, the voltage at the non-inverting pin is -0.7, and the situation is the aforementioned.

What I understand about potential difference

The basic electronic says a source voltage is the result of a potential difference between its poles.

It is more practical to consider a voltage source as an element having electric potential energy, which is manifested through a potential difference between its terminals, and has the ability to move electric charge.

With that I understand that a battery of 1.5V has a positive terminal with 1.5 volt charge and a negative terminal to 0 volts of charge.

The Volt is the unit used to measure electric potential or potential difference. To measure amount of electric charge, the Coulomb is used.
The fact that a battery is 1.5 volts, means that the potential difference across its terminals is 1.5 volts, and is a measure of the battery capacity to do the work of moving electric charge, or that is, generate an electric current.

What I understand about diode

A diode is a semiconductor electronic component divided into two poles, one negative and one positive so with a potential difference between them; It is separated by a resistive layer between the poles.

The diode is not divided into two poles, but it consists of semiconductor material N type and P type These materials are obtained from pure silicon, by adding impurities (doping), which provide electrons (N type) or generate holes (P type).
This device doesn't presents a potential difference between its terminals, ie, if we connect a voltmeter to the terminals, there will not measure voltage. The associated quantum effect is the Potential Barrier. Basically this is a distribution of the electric potential energy in the crystal lattice formed by silicon, which acts as a barrier that prevents the passage of electrical charges, whether these holes or electrons.

When a diode is connected to battery directly (negative diode with negative source) etc, free electrons on the negative side of the battery repel the electrons free from the negative side of the diode to destroy the resistive layer and the current flow through the circuit.

The connection that you describe, is called forward bias diode. Nothing in the internal diode is "destroyed". What happens is that the potential energy of the battery is greater than the energy implicit in the potential barrier, which, it is reduced, allowing the passage of electrical charges.

What I can not understand

Let's say I have a source with 0 volts in its negative pole and a diode with -7 volts in its negative pole. Logically, the electrons on the negative side of the diode should go to the negative side of the source and thereby increase the depletion layer in the diode even when polarized directly to the source.

I guess you're talking about a 0 volt battery with the positive terminal and -7 volts on the negative terminal. In this case the only thing that interests us is that the difference in potential corresponding to this battery is 7 volts. Then in the application with the diode, the only thing we care about is whether it is forward bias or reverse bias.
Charges of equal polarity repel, so free in the N type materials electrons are repelled when connected to the lower potential pole. In this case, it is forward biased. Different polarity charges attract, so that the free electrons in the N-type material are attracted when connected to the pole of higher potential battery. In this case it is reverse biased.