The ratio of mutual flux to primary flux is the same as the coefficient of coupling. However, the flux linkage of the secondary is the mutual flux multiplied by the number of turns of the secondary. Therefore the flux linkage ratio is not the same as the coefficient of coupling and may have a value greater than 1 unlike the coefficient of coupling which is always less than 1 and greater than 0. Lets prove that out from some known laws.
So from Faraday's law we can derive the following relationship between the mutual inductance and the mutual flux:
\$ L_m = N_s \cdot \frac{\delta \phi_m}{\delta i_p} \$
the mutual flux is the portion of the flux from the primary that overlaps with the secondary. Therefore the following is always true:
\$ \phi_m \leq \phi_p\$
We also know flux as a function of current can be derived from the following equation:
\$ L = \frac{\phi \cdot N}{i} \$
Solving for the flux we get:
\$ \phi = \frac{L \cdot i}{N}\$
Lets also define the ratio of flux linkage with the secondary coil to primary coils flux:
\$ R_\lambda = \frac{\lambda_s}{\phi_p} \$
Solving for the flux linkage and substituting the primary flux with the earlier equation we get:
\$ \lambda_s = R_\lambda \cdot \phi_p \$
\$ \lambda_s = R_\lambda \cdot \frac{L_p \cdot i_p}{N_p} \$
\$ \lambda_s = \frac{R_\lambda \cdot L_p}{N_p} \cdot i_p \$
It is important to note here that the flux linkage, \$\lambda_s\$, is the mutual flux multiplied by the number of turns of the secondary winding:
\$\lambda_s = N_s \cdot \phi_m\$
Alternatively:
\$\phi_m = \frac{\lambda_s}{N_s}\$
Since we know that flux linkage of the secondary is the portion of the primaries flux which is mutual multiplied by the number of turns in the secondary, then it stands to reason the ratio between the flux linkage and the primary flux can be any positive value, including values greater than 1.
Next we can take the derivative of our first equation by substitution:
\$ L_m = N_s \cdot \frac{\delta}{\delta i_p} (\frac{\lambda_s}{N_s}) \$
\$ L_m = N_s \cdot \frac{\delta}{\delta i_p} (\frac{\frac{R_\lambda \cdot L_p}{N_p} \cdot i_p}{N_s}) \$
\$ L_m = N_s \cdot \frac{\delta}{\delta i_p} (\frac{R_\lambda \cdot L_p}{N_p \cdot N_s} \cdot i_p) \$
\$ L_m = N_s \cdot \frac{R_\lambda \cdot L_p}{N_p \cdot N_s} \$
\$ L_m = \frac{R_\lambda \cdot L_p}{N_p} \$
We also know the relationship between mutual inductance and the coefficient of coupling with this equation:
\$ L_m = k \cdot \sqrt{L_p \cdot L_s} \$
We can now substitute this in to get an equation showing the relationship between the flux link ratio and the coefficient of coupling:
\$ k \cdot \sqrt{L_p \cdot L_s} = \frac{R_\lambda \cdot L_p}{N_p} \$
Now we can solve for the flux linking ratio and simplify:
\$ k \cdot \sqrt{L_p \cdot L_s} = R_\lambda \cdot \frac{L_p}{N_p} \$
\$ k \cdot \sqrt{L_p \cdot L_s} \cdot \frac{N_p}{L_p} = R_\lambda\$
\$ k \cdot \frac{N_p \cdot \sqrt{L_p \cdot L_s}}{L_p} = R_\lambda\$
Since all of the constants here are positive values we can further reduce this to:
\$ k \cdot N_p \cdot \sqrt{\frac{L_s}{L_p}} = R_\lambda\$
It should now be clear that the flux link percentage (\$R_\lambda\$) is not the same value as the Coefficient of Coupling (\$k\$).
While most of the derivations above I did myself you can find most of the equations I pulled in as laws (Like Faraday's Law in the first equation) in this wonderful online book that is worth taking a read for a deeper explanation of the subject.
