I'm trying to find some helpful intuition as to why the lagrangian in classical mechanics has the form $$L = T - V.$$
The Lagrangian does not have this form in all cases. It has this form in the case of conservative forces that can be derived from a potential, like:
$$
\vec F = - \frac{\partial V}{\partial \vec r}\;,\tag{1}
$$
where the $\vec r$ are the coordinates.
See that negative sign there in Eq. (1)? There is no way to have any a priori "intuition" regarding that sign, since it is a matter of convention. Please keep that in mind.
So far, all the explanations that I've found resort to motivating this definition by showing that it correctly reproduces $F = ma$ after finding the corresponding Euler-Lagrange equations.
This is not the explanation that I have seen. I prefer an explanation that starts from Newtonian equations of motions for each particle $i$
$$
\vec F_i = m_i \ddot{\vec r_i}\;,
$$
in cartesian coordinates $\vec r_i = (x_i, y_i, z_i)$,
introduces the definition of kinetic energy
$$
T = \frac{1}{2}\sum_{i=1}^N m_i {|\dot{\vec r_i}|}^2\;,
$$
and then arrives at the Lagrange equations:
$$
\frac{d}{dt}\frac{\partial T}{\partial \dot q_j} - \frac{\partial T}{\partial q_j} = Q_j\;,\tag{2}
$$
where the $q_j$ are generalized coordinates, and $Q_j$ is the $j$th component of the generalized force:
$$
Q_j = \sum_i \vec {F_i}\cdot\frac{\partial \vec r_i}{\partial q_j}\;.
$$
(See Whittaker's textbook "Analytical Dynamics" for a derivation.)
When the generalized forces can be written in the form:
$$
Q_j = -\frac{\partial V}{\partial q_j}\;,
$$
where $V=V(q)$, such forces are called "conservative" and the Lagrange equations of Eq. (2) become:
$$
\frac{d}{dt}\frac{\partial T}{\partial \dot q_j} - \frac{\partial T}{\partial q_j} = -\frac{\partial V}{\partial q_j}\;,\tag{3}
$$
Or, rearranging Eq. (3), we find
$$
\frac{d}{dt}\frac{\partial (T)}{\partial \dot q_j} - \frac{\partial (T - V)}{\partial q_j} = 0\;.\tag{4}
$$
or
$$
\frac{d}{dt}\frac{\partial (T - V)}{\partial \dot q_j} - \frac{\partial (T - V)}{\partial q_j} = 0\;.\tag{5}
$$
or
$$
\frac{d}{dt}\frac{\partial L}{\partial \dot q_j} - \frac{\partial L}{\partial q_j} = 0\;.\tag{6}
$$
Eq. (6) is ultimately the reason why $L = T - V$ is useful. The reason for the minus sign in $L = T - V$ ultimately results from the convention we use to determine the force from the potential:
$$
Q_j = -\frac{\partial V}{\partial q_j}
$$
I don't find myself satisfied after reading these explanations, maybe because I find them too circular: we define the Lagrangian such that it spits the equations of motion that we know are correct.
The reasoning is not circular. Being consistent is not the same as being circular.
In addition, the reason the Lagrange formulation is useful is not just because it is consistent with Newton's equations, but because it is easier to use in many cases.
Personally, I believe that the beauty of Lagrangian mechanics comes from formulating physical law in terms of a principle of least (or generally extremal) action.
Ok.
I love the intuition of particles following the path that requires the least "effort".
This does not seem to have much to do with "intuition" unless you can somehow intuit the potential.
For this intuition to be complete, I need a motivation as to why we can wrap this concept of "effort that a particle requires to reach some position" in the time integral of $T - V$.
I don't see how this can be intuitive, in any real sense, other than an intuition derived from solving lots of physics problems.
