Here are two separate ways to address the issue you bring up. One is more mathematical---comparing the relations $mv$ and $\frac{1}{2}mv^2$. The other has more to do with force and energy, which I'm calling physical.
Mathematical
Let's imagine two objects that are moving in the same direction collide with each other. Just to keep things simple, let's also imagine that they move in the same direction after the collision. (This can always be set up, so you don't lose anything by assuming it.)
Before and after the collision, the quantity
$$p_\text{tot} \equiv m_1v_1 + m_2v_2 \tag{1}$$
is unchanged. The speeds may have changed from before & after the collision, but you can plug in either set (either the initial speeds or the final speeds) that sum won't change.
Now what can be said about the quantity
$$2K_\text{tot} \equiv m_1v_1^2 + m_2v_2^2?\tag{2}$$
(I moved the $\frac{1}{2}$ to the other side; hope that's okay with you. Just makes the expression look more similar.) Well, not much really. They are both composed of the same quantities, but they aren't necessarily the same because there's no mathematical way to manipulate Eqn. 1 to make it look like Eqn. 2. Try it, you won't be able to. Here's what I mean. I can multiply $p_\text{tot}$ by $v_{1f}$ (that's object 1's final velocity) and end up with an invented quantity I'm calling $Q$:
$$Q\equiv p_\text{tot}v_{1f} = m_1v_1v_{1f} + m_2v_2v_{1f}. \tag{3}$$
Now that quantity is the same before and after the collision. How do I know? Because $p_\text{tot}$ is the same, so $p_\text{tot}$ multiplied by the same number $v_{1f}$ must also be the same.
That's what I mean when I said you can't manipulate $p_\text{tot}$ to make it look like kinetic energy. So there's no reason kinetic energy should be the same before and after the collision.
Physical
The momentum of a system of objects is the same before and after the collision if the net impulse on the system is zero:
$$\int F_\text{net}\,dt = \Delta p$$
That is Newton's 2nd law, but written in a different form than you might have seen.
So now we know when and "why" momentum is constant. What about kinetic energy? That's actually harder. The governing equation is
$$\sum_i \vec F_i\cdot d\vec s = \Delta K + \Delta U + \Delta E_\text{thermal} + \cdots$$
In other words, the sum of the external works on your system equals the change in total energy, but that doesn't tell you anything about the kinetic energy. Energy can change forms. So if kinetic energy is lost in some collision, it went into potential, thermal, etc.