It is because pound-feet correspond to torque, but not only torque. Energy has the same physical dimension of $\mathrm{kg \cdot m^2 \cdot s^{-2}}$ or $\mathrm{lbs \cdot ft^2\cdot s^{-2}}$, yet they are not directly comparable.
In case of torque this is the result of a cross product of two vectors $\vec{r} \times \vec{F}$ (length of the lever arm multiplied by the force); while with energy this is a dot product $\vec{r} \cdot \vec{F}$ (force multiplied by the distance over which it acted, while also considering the cosine of the angle between them).
Another bad habit that can be seen here is the implicit assumption of standard gravity $g$. This used to be common with SI units and "kilogram-force" $\mathrm{kg_F}$ or kiloponds $\mathrm{kp}$, but is now seen very rarely. However, with Imperial and US customary units this usage is still fairly common, "pounds" are often used both as a unit of mass, and as a unit of force, written as $\mathrm{lbf}$ or $\mathrm{lb_f}$.
So a "pound foot" could, in different context, also mean a pound-foot of energy – that is, the energy expended in moving one pound (of force) over a distance of one foot, while implicitly assuming standard gravity $g_0$. However, historically this was solved in English by changing the order and calling this unit a foot-pound instead. The unit can still be encountered in various texts, for instance when describing muzzle energies of small arms.