In this answer I go specifically into the assertion that the Lagrangian for the free particle must be an expression that is quadratic in velocity.

The presentation in L&L isn't so much a derivation, but a consistency argument.

The demands are:
A quantity that is:
-independent of orientation
-consistent with galilean relativity

Momentum $m\vec{v}$ has the property that for any inertial coordinate system momentum is a conserved quantity (hence consistency with galilean relativity). But momentum is vectorial, of course, so it doesn't satisfy the independent-of-orientation demand.

For the purpose of addressing the implications of a quadratic expression for quantity of motion:

If we have two masses, $m_1$ and $m_2$, then the straightforward thing to do is to express their respective velocities with respect to some inertial coordinate system. However, it is helpful to use the Common Center of Mass (CCM) of the two masses.

Definitions:
$V_r$ $\Rightarrow$ the relative velocity between $m_1$ and $m_2$
$V_c$ $\Rightarrow$ the velocity of the CCM relative to some chosen origin

$v_1$ velocity of mass 1 relative to the CCM
$v_2$ velocity of mass 2 relative to the CCM

Then we have:

$$ v_1 = V_r\frac{m_2}{m_1 + m_2} \tag{1} $$ $$ v_2 = -V_r\frac{m_1}{m_1 + m_2} \tag{2} $$

Reminder: here $v_1$ and $v_2$ are defined as velocity relative to the Common Center of Mass.
Relative to the CCM: $m_1v_1 = m_2v_2$
Which rearranges to: $\frac{m_1}{v_2} = \frac{m_2}{v_1}$

This notation in term of velocity relative to the CCM embodies galilean relativity.

The notation (1) and (2) allows a way of evaluating what an quadratic expression for quantity of motion will do.

As quadratic expression for quantity of motion we take: $mv^2$

Then the combined quantity of motion that we attribute to $m_1$ and $m_2$ is as follows:

$$ m_1\left(V_c + V_r\frac{m_2}{m_1 + m_2} \right)^2 + m_2\left(V_c - V_r\frac{m_1}{m_1 + m_2} \right)^2 \tag{3} $$

When you develop that expression quite few terms drop away against each other. The end result is as follows:

$$ (m_1 + m_2) V_c^2 + \frac{m_1m_2}{m_1 + m_2} V_r^2 \tag{4} $$

Next we consider what (4) implies for the outcome of a perfectly elastic collision.

After the collision the minus sign is with the other velocity. The squaring eliminates that minus sign; other terms with the minus sign drop away, and the expression for the motion after the elastic collision comes out the same as (4)

This informs us that the quantity $mv^2$ has the following property: the total amount of quantity of motion $mv^2$ that we attribute is the same before and after an elastic collision, for any inertial coordinate system; consistency with galilean relativity.

Another way of looking at (4):
With a perfectly inelastic collision: for any inertial coordinate system the expresssion for the amount of quantity of motion $mv^2$ that is lost in an inelastic collision is the same: consistency with galilean relativity.

Discussion:

You wouldn't necessarily expect that an expression for quantity of motion that is quadratic is consistent with galilean relativity, but indeed we get that consistency.

It's quite a marvelous property: the squaring eliminates directional information, so the quantity is independent of orientation.

Now the question: the modern notion of quadratically expressed quantity of motion, kinetic energy, is defined as $\tfrac{1}{2}mv^2$
Where does that factor 1/2 come from?

To motivate the introduction of the concept of kinetic energy we derive the work-energy theorem. If $F=ma$ is granted as axiom then the work-energy relation follows as theorem.

The starting point:

$$ F = ma \tag{5} $$

The next step is to evaluate an integral: integrate both sides with respect to the position coordinate, integrating from starting point $s_0$ to final point $s$

$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \tag{6} $$

The work-energy theorem hinges on the fact that the two factors in the right hand side of (2), acceleration $a$ and position coordinate $s$, are not independent of each other; acceleration is the second derivative of position.

We proceed to develop the right hand side of (6).

In the steps starting with (9) the integrand $a$ is converted to velocity and the differential $ds$ is converted to $dv$, using the relations (7) and (8)

$$ v = \frac{ds}{dt} \ \Leftrightarrow \ ds = v \ dt \tag{7} $$

$$ a = \frac{dv}{dt} \ \Leftrightarrow \ dv = a \ dt \tag{8} $$

(9) corresponds to the right hand side of (6)

$$ \int_{s_0}^s a \ ds \tag{9} $$ $$ \int_{t_0}^t a \ v \ dt \tag{10} $$ $$ \int_{t_0}^t v \ a \ dt \tag{11} $$ $$ \int_{v_0}^v v \ dv \tag{12} $$

First (7) was used to change the differential from $ds$ to $dt$, with corresponding change of limits. Next (8) was used - with change of limits - to arrive at (12).

So we have the following mathematical relation:

$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{13} $$

Notice especially:
The definitions (7) and (8), in combined form, are sufficient to imply (13).
Stated differently: (13) expresses: whenever you have a quantity $s$, its first derivative, and its second derivative, an expression with the form of (13) is valid. That is: (13) is a mathematical property, not tied to any particular physics context.

We use (13) to go from (6) to the work-energy theorem:

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{14} $$

When a quadratic expression has a factor of 1/2 in front of it: that factor 1/2 is a signature of integration.

To motivate the concept of kinetic energy I prefer the derivation of the work-energy theorem. The work-energy theorem acts as a hub, tying everything together.

Discussion:

It is not common for physiscs textbook authors to mention that the concept of kinetic energy slots in with galilean relativity. Some authors may not even be aware of the significance of that.

My guess is that Landau was prompted by an intuition that it is important and worthwhile to demonstrate that a quadratic expression for quantity of motion, $mv^2$, slots in with galilean relativity.