Why can we get maximum work from a reversible process?

This is for a heat engine. The maximum work corresponds to the generation of the minimum change in entropy, and the minimum change in entropy is zero. A reversible process has zero change in entropy. Therefore a reversible process does maximum work.

The work done on a heat engine is $$W=Q_c-(-Q_h)$$ where $Q_c$ is the heat entering the cold reservoir and $Q_h$ is the heat entering the hot reservoir. Given that $Q=T Delta S$ then after a little algebra we get that the efficiency is $$eta = 1-frac{T_c Delta S_c}{-T_h Delta S_h}$$ Since for a reversible process $-Delta S_h = Delta S_c$ and for irreversible processes $-Delta S_h < Delta S_c$ we see that the maximum efficiency is for a reversible process.

If you use energy from a system to compress a spring then all of this energy can be recovered by making the spring to do work as it expands back to its original length. If you use energy from a system to compress an ideal gas then all of this energy can be recovered by making the gas do work as it expands back to its original volume. If you use energy from a system to raise a weight then all of this energy can be recovered by making the weight do work as it falls back to its original height. These are all examples of reversible processes.

If you use energy from a system to overcome friction then this energy cannot all be recovered again to do work. This is because friction is an irreversible process. Some of the energy used to overcome friction can be recovered in a heat engine, but this will have an efficiency that is less than $1$, so some energy is simply lost as heat and cannot be used to do work. The difference between reversible and irreversible processes is that energy used in an irreversible processes cannot all be recovered to do work.

conclude that the work done by friction or any opposing force in both the scenarios will be the same, then how can we say that we can get maximum work from a reversible process? Shouldn't both reversible and non-reversible process give the same result?

To address this, we introduce a concept known as entropy production (see here), so if have viscous forces or such, immediately we are speaking of an irreversible process which involves a lesser heat transfer into the system and more wastage. See here for the intuition behind it.

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You can actually derive it from the Clausius inequality, first of all, consider two processes: Process-A and process-B, which ends up in the same final state. One is reversible and the other is irreversible both involving some infinitesimal heat and infinitesimal work moving from state-1 $to$ state-2. Writing down the Clausius inequality for process-B:

$$ dS_{ 1 to 2} geq frac{dq}{T}$$

For this process, we can write the entropy using the reversible process:

$$ dS_{1 to 2} = frac{dq_{rev} }{T}$$

Pluggin this, we reach:

$$ dq_{rev} geq dq$$

Or,

$$ dq_{rev} - dq geq 0 tag{1}$$

Now, head back to the second law of thermodynamics:

$$ dU_{1 to 2} = dq + dw = dq_{rev} + dw_{rev} tag{2}$$

Since, quantities in thermodynamics are path independent , we can say that sum of infinitesimal heat transfer plus the infinitesimal work done should be same for both processes.

Rearranging eq-(2),

$$ dq_{rev} -dq= dw - dw_{rev}$$

Using (1) and (3),

$$dq_{rev} - dq = dw-dw_{rev} geq 0$$

Or,

$$ dw geq dw_{rev}$$

In the sign convention I'm using, work is negative when energy leaves the system and hence to make work done by system as positive, we multiply by a minus:

$$ - dw_{rev} geq -dw$$

taking the modulus:

$$ |dw_{rev}| geq | dw$$

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Inspired from Atkin's physical chemistry, see page-81 bottom-most paragraph under section "The Clausius inequality"