According to the Rankine–Hugoniot equation, the pressure P2 after the shock wave passes is $$\frac{P_{2}}{P_{1}}=\frac{2 \gamma M_{1}^{2}}{\gamma+1}-\frac{\gamma-1}{\gamma+1}$$
Does this indicate that when the shock wave passes, the pressure difference causes the object to blown off? I want to know the force that an object receives when a shock wave passes.
A pressure is, in terms of units, just a force per unit area. In principle, one may think that you need only multiply the pressure by the area on which it impinges at normal incidence to get a force but that will not address your question.
In the case of a shock, the relevant thing to think about is a force density, i.e., the gradient in the pressure. Then you integrate over the affected volume to get the net force acting on that volume.
For shock waves in Earth's atmosphere under normal conditions near sea level, the gradient scale length of an atmospheric shock wave is roughly a micron, i.e., roughly the mean free path for particle-particle collisions. Also in Earth's atmosphere, it is typically safe to assume the gas is composed of diatomic molecules, thus the equation of state can be expressed in terms a polytrope index $\gamma$ ~ 5/3.
Thus, the ratio fo pressures goes to: $$ \frac{ P_{2} }{ P_{1} } = \frac{ 5 \ M^{2} - 1 }{ 4 } \tag{1} $$
Suppose we have a strong shock with M = 5, then the numerical value of the pressure ratio from Equation 1 will be ~31.
The pressure gradient can be approximated as: $$ \nabla P \sim \frac{ P_{2} - P_{1} }{ L } = \frac{ 5 \ P_{1} }{ L } \left[ \frac{ M^{2} - 1 }{ 4 } \right] \tag{2} $$
If we assume a planar surface and since pressure acts at a normal incidence, integrating over the volume will reduce to a 1D integral of the form: $$ \mathbf{F} = \int_{V} \ d^{3}x \ \nabla P \sim \int_{S} \ dA \ \int_{0}^{x_{o}} \ dx \ \nabla P \tag{3} $$
Does this indicate that when the shock wave passes, the pressure difference causes the object to blown off? I want to know the force that an object receives when a shock wave passes.
In the simplest scenario where the outward unit normal vector of the shock is parallel to the unit normal vector of the object on which the pressure wave is acting, then one can make things incredibly simple and reduce Equations 2 and 3 down to: $$ F \propto A \ 5 \ P_{1} \left[ \frac{ M^{2} - 1 }{ 4 } \right] \tag{4} $$ That is, the force is as simple as the pressure times the area.
For more realistic objects and non-parallel incidence, things will obviously change.
The pressure differential across the shock wave is what blows objects apart. Multiply this pressure difference by the area in question to get the instantaneous force involved.
