What is wrong with this approach?
TL;DR You are correct that the force needed to maintain constant velocity of the cart is
$$F = \frac{dp}{dt} = \frac{d}{dt} m v = m \frac{dv}{dt} + v \frac{dm}{dt} = v \frac{dm}{dt} \qquad \text{where} \qquad \frac{dv}{dt} = 0 \tag 1$$
which follows directly from the second Newton's law of motion. However, be extra carefulbeware with how you apply the procedure from theproduct rule for $\frac{d}{dt}(mv)$ as done in Eq. (1) because it does not apply in general to variable-mass systems. It works only in a reference frame in which sand has horizontal momentum equal to zero, which is the case in your example. Otherwise, the only way to solve this problem would be via momentum. Check the "Variable-mass system" Wiki article for more details.
The key to understanding the "contradiction" you mention is that kinetic energy is lost to the perfect inelastic collision. I show below how to calculate the lost kinetic energy in two ways: (i) via conservation of energy, and (ii) via conservation of momentum. This explains why the force you got from the conservation of energy is half the force from Eq. (1).
Lost kinetic energy via conservation of energy
The work $dW$ done by the force $F$ over displacement $dx$ is
$$dW = F \cdot dx = v \frac{dm}{dt} dx = v \frac{dx}{dt} dm = v^2 \cdot dm$$
Integrating the above equation we get that the total work done by the force $F$ to give sand of mass $M$ required velocity $v$ is
$$W = M v^2$$
It does not matter whether the particle gained mass or velocity, the kinetic energy is always $K = \frac{1}{2} m v^2$. The change in kinetic energy in your example is
$$\Delta K = \frac{1}{2} (m + M) v^2 - \frac{1}{2} m v^2 = \frac{1}{2} M v^2$$
where $m$ is mass of the empty cart. From the work-energy theorem and law of conservation of energy, kinetic energy lost to the perfect inelastic collision is
$$\boxed{K_\text{lost} = W - \Delta K = \frac{1}{2} M v^2}$$
Lost kinetic energy via conservation of momentum
Conservation of momentum for perfect inelastic collision is
$$(m \cdot v) + (dm \cdot 0) = (m + dm) v'$$
where $m$ is mass of the cart plus sand added up until some point in time, $v$ is the cart velocity maintained at constant value by applying force from Eq. (1), $dm$ is additional infinitesimally small mass of sand we put into the cart, and $v'$ is new velocity due to the perfect inelastic collision
$$v' = \frac{m}{m + dm} v$$
The infinitesimally small loss of kinetic energy due to the infinitesimally small added mass of sand $dm$ is
$$dK = \frac{1}{2} m v^2 - \frac{1}{2} (m + dm) v'^2 = \frac{1}{2} m v^2 \frac{dm}{m + dm}$$
By rearranging the above equation we get
$$m \cdot dK + \underbrace{dm \cdot dK}_{\approx 0} = \frac{1}{2} m v^2 \cdot dm$$
We can neglect the product of two infinitesimally small values ($dm \cdot dK \approx 0$) which gives
$$dK = \frac{1}{2} v^2 \cdot dm$$
By integrating the above equation we finally get that the total kinetic energy lost in the perfect inelastic collision is
$$\boxed{K_\text{lost} = \frac{1}{2} M v^2}$$
where $M$ is total mass of sand in the cart.