How to Find Thevenin's Voltage & Resistance?
Thevenin's theory states that for a circuit consisting purely of fixed sources we can replace part of the circuit with a voltage source \$ (V_{th}) \$ and a series impedance \$ (R_{th}) \$ to have the same result.
To do this we first remove our load and calculate \$ (V_{th}) \$ as the open circuit voltage you would see across the load.  Next we replace any sources with their output impedance (short circuit for voltage source, open circuit for current source) and calculate the impedance you would measure.
Taking your circuit as an example
$$V_{th} = 7 + (28-7) \cdot \dfrac{1}{4+1} = 11.2 \text{ volt
}$$
$$ R_{th} = 4 || 1 = \dfrac{4 \times 1}{4+1} = 0.8 \text{ ohm}$$
We can now put our load back into the simplified circuit and from the loads point of view the circuit is equivalent.
The best way to verify this for yourself is to assume you have an output voltage between 0V and 11.2V (depending on load). Calculate the current in the load resistor for each voltage and the resistor value.  Do this for both the original circuit and the Thevenin equivalent.  Both sets of answers should be identical.

For a voltage source you replace it with a short because that is its output impedance.  You can draw as much current as you want and the voltage does not change.
For a current source you replace it with an open circuit because that is its output impedance.  Current does not change irrespective of voltage.
If you don't remove the load then you may not have simplified the circuit depending on how you create \$ V_{th} \$.