$S_2$ is a solenoid with more radius than $S_1$, the magnetic field of the solenoid $S_2$ will be along the axis and constant, so the magnetic force on an element of solenoid $S_1$ will be,

$d{\overrightarrow{F}}_1=i\left(\overrightarrow{l}\times \overrightarrow{B}\right)$

$d{\overrightarrow{F}}_1=ilB\left(\hat{\mathrm{j}}\times \left(-\hat{\mathrm{i}}\right)\right)=iBl\hat{\mathrm{k}}$

Similarly, for just next coil the magnetic field is in same direction, but the current element is in opposite direction, so,

$d{\overrightarrow{F}}_1=ilB\left(-\hat{\mathrm{j}}\times \left(-\hat{\mathrm{i}}\right)\right)=-iBl\hat{\mathrm{k}}$.

From Symmetry, we can say that the overall force will be cancelled to the whole coil, hence, ${\overrightarrow{F}}_1=0$

Similarly, from coil $S_1$ to $S_2$, The magnetic field of the inner solenoid is symmetric about its axis, but not exactly along the axis of the outer solenoid. As the outer solenoid is placed symmetrically the push of any coil will be equal and opposite to the pull of coil placed exactly the opposite distance from the centre. Hence, the net force on the outer coil will be zero, i.e., ${\overrightarrow{F}}_2=0$.