I am primarily looking for the difference in definitions to see how they differ. (Emphasis in original).
The kinetic energy $T$ is defined as:
$$
T = \sum_{i=1}^N \frac{1}{2}m_i \vec v_i^2
$$
$$
= \sum_{i=1}^N \frac{1}{2}m_i\left(\dot x_i^2 + \dot y_i^2 +\dot z_i^2\right)\;,\tag{1}
$$
where $x$, $y$, and $z$ are a fixed set of rectangular/cartesian coordinate axes (as you might expect), not generalized coordinates.
For a conservative force field $\vec F(\vec x)$, the single-partial potential energy $U$ is usually defined (up to a constant) via:
$$
\vec F =-\vec \nabla U\;.
$$
By analogy, I define $V = V(\vec q_1, \vec q_2,\ldots)$ via the force on particle i:
$$
\vec F^{(i)} = -\vec \nabla_i V\;.
$$
For example, if the system of particles is non-interacting, other than via a single-particle potential $U$, then we can write $V = \sum_i U(\vec x_i)$. (Note that even $V=V(q_1,q_2,\ldots)$ is not exactly the most general form of potential, but this is explained further in an addendum.)
The "total mechanical energy" $E_{TM}$ is defined as:
$$
E_{TM} = T + V\;,
$$
but be careful, because this thing I'm calling "$E_{TM}$" might also be called the "mechanical energy" or the "total energy" or the "energy." I'm hanging a couple subscripts off of my $E$ symbol to indicate exactly what I mean, but no one else will ever do that.
The "Lagrangian energy function," as you have called it, is defined properly as you have defined it in terms of generalized coordinates as:
$$
h = \sum_{i=1}^{3N}\dot q_i\frac{\partial L}{\partial \dot q_i} - L\;,
$$
but be careful, because someone might also call this the "mechanical energy" or the "total energy" or the "energy."
And, of course, L is defined as:
$$
L = T - V\;.
$$
Now, you have the proper definitions, you should be able to figure out for yourself when the "Lagrangian energy function" is equal to the "total mechanical energy."
For some further assistance, see my answer here, and see below, and perhaps consult a graduate textbook on Classical Mechanics, for example, the textbook by Whittaker (some of which is rewritten below).
Addendum:
You might be interested to know that the Lagrangian equations of motion in the form:
$$
\frac{d}{dt}\frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = 0\;,
$$
is not the most general form.
Supposing that the force can not be written as the gradient of a potential, we can still write a form of the Lagrange equations of motion as:
$$
\frac{d}{dt}\frac{\partial T}{\partial \dot q_i} - \frac{\partial T}{\partial q_i} = Q_i\;,
$$
where the $Q_i$ is a generalized force, which is defined via the work done as the generalized coordinate $q_i$ changes by an infinitesimal amount: $dW_i = Q_i(q_1,q_2,\ldots) dq_i$ (no sum on i implied).
In the case where we can define a potential energy function
$$
V(q_1, q_2, \ldots)\;,
$$
then we can write:
$$
\frac{d}{dt}\frac{\partial T}{\partial \dot q_i} - \frac{\partial T}{\partial q_i} = -\frac{\partial V}{\partial q_i}\;,
$$
and then we can recover the other form of the equations of motion.
Addendum 2
So, when are the "total mechanical energy" ($E_{TM}$) and the "Lagrangian energy function" ($h$) equal?
To answer this we should consider a slight generalization of the potential energy $V$. But at first, a let's consider a potential that only depends on the $q_i$.
Previously, I indicated that if we can find a potential $V$ such that:
$$
-\frac{\partial V}{\partial q_i} = Q_i\;,
$$
then we recover the usual Lagrangian equations of motion:
$$
\frac{d}{dt}\frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = 0\;.
$$
From the above equation of motion we can show that $h=E_{TM}$ whenever:
$$
\dot q_i\frac{\partial T}{\partial \dot q_i} = 2T\;. \tag{A}
$$
Eq. (A) is the condition that the "total mechanical energy" ($E_{TM}$) and the "Lagrangian energy function" ($h$) are equal, given that we can write $Q_i = -\frac{\partial V}{\partial q_i}$.
So, what is the generalization of this result to velocity dependent potentials?
Suppose that we want to be a little more general and write
$$
V=V(q_1,q_2,\ldots,\dot q_1, \dot q_2,\ldots)
$$
where now we define V via:
$$
Q_i = -\frac{\partial V}{\partial q_i} + \frac{d}{dt}\frac{\partial V}{\partial \dot q_i}\;.
$$
In this case, the "total mechanical energy" ($E_{TM}$) and the "Lagrangian energy function" ($h$) are equal whenever:
$$
\dot q_i\frac{\partial T}{\partial \dot q_i}-\dot q_i\frac{\partial V}{\partial \dot q_i} = 2T\;. \tag{B}
$$
Condition (B) reduces to condition (A) when the potential is velocity independent.
Addendum 3
OP asks a question about coordinate transformations in the comments. To answer this question, it is easiest to return to the usual case where $Q_i=-\frac{\partial V}{\partial q_i}$ and where Condition (A) then holds.
The condition that $h=E_{TM}$ is then:
$$
\dot q_i\frac{\partial T}{\partial \dot q_i} = 2T\;.
$$
In terms of coordinate transformations, first consider the familiar case when the coordinate transformation don't depend on velocity or time explicitly:
$$
x_i = x_i(q_1,q_2,\ldots)\;.
$$
$$
y_i = y_i(q_1,q_2,\ldots)\;.
$$
$$
z_i = z_i(q_1,q_2,\ldots)\;.
$$
In this case, it is straightforward to show using Eq. (1) above and using
$$
\frac{\partial \dot x_i}{\partial \dot q_j} = \frac{\partial x_i}{\partial q_j}
$$
that
$$
\dot q_i\frac{\partial T}{\partial \dot q_i} = 2T\;.
$$
And so, in this case, $h=E_{TM}$.
But, in general, when the $x_i$, $y_i$, and $z_i$ depend on the $\dot q_j$ as well, we instead find that:
$$
h = E_{TM} + \sum_j m^{(j)}\dot x_j\left[\dot x_j - \sum_i\dot q_i\frac{\partial \dot x_j}{\partial \dot q_i}\right]
+ \sum_j m^{(j)}\dot y_j\left[\dot y_j - \sum_i\dot q_i\frac{\partial \dot y_j}{\partial \dot q_i}\right]
+ \sum_j m^{(j)}\dot z_j\left[\dot z_j - \sum_i\dot q_i\frac{\partial \dot z_j}{\partial \dot q_i}\right]\;.\tag{2}
$$
The quantities in the square brackets in Eq. (2) above are zero whenever the coordinate transformations do not explicitly depend on the velocities.
