Voltage Divider returns incorrect voltage reading
The answer is simple. Your real voltmeter is not ideal thus, it will have an internal resistance around \$1M\Omega\$. A better voltmeter will have internal resistance around \$10M\Omega\$
So your circuit will look like this:



simulate this circuit &ndash; Schematic created using CircuitLab
And the output voltage will be :
$$V_O = 7.88V \times \frac{560k\Omega||1M\Omega}{4.7M\Omega +{560k\Omega||1M\Omega} } \approx 0.56V$$