For massive spinors "right-handed" and "left-handed" chirality isn't tied so much to true rotations, as to the casting of Lorentz transformations as "space-time rotations". In this case, a very popular short answer to the conceptual question is that Lorentz transformations "rotate" $(1/2, 0)$-spinors one way in space-time, and $(0, 1/2)$-spinors the opposite way, while the space inversion corresponding to the parity transformation "rotates" one type of spinor into the other.
But the correct understanding of spinor chirality, and its connection to parity, is closely related to another pair of very familiar concepts, the contravariance and covariance duality of 4-vectors. Recall that in Minkowski space-time a covariant vector $v_\mu$ is the space-inverted of its contravariant counterpart $v^\mu$, and so the parity transformation is the metric itself. The more confusing issue is that, regardless of this connection, each irrep induces its own distinct set of contravariant and covariant spinors, the two constructions being related by a complex conjugation that in a quantum setting compounds the usual role of complex conjugation in defining bra-ket duality.
Now for some details:
Why the "right-handed" and "left-handed" monikers:
Say a given space-time Lorentz transformation, for a rotation of angle $\theta$ around axis $\vec{n}_\theta$, $\vec{\theta} = \theta\vec{n}_\theta$, and a boost of rapidity $\zeta$ in direction $\vec{n}_\zeta$, $\vec{\zeta} = \zeta\vec{n}_\zeta$, reads
$$
L = e^{\vec{\theta}\cdot\vec{J} + \vec{\zeta}\cdot \vec{K}}
$$
where $\vec{J}$, $\vec{K}$ are the usual rotation and boost generators in Minkowski space-time. Then its equivalents on the (1/2,0) and (0,1/2) spinor reps are
$$
\Lambda_{(1/2,0)} = e^{\vec{w}\cdot\vec{\sigma}/2} \equiv e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2}, \;\;\;\;\;\Lambda_{(0,1/2)} = \left(\Lambda_{(1/2,0)}^{-1}\right)^\dagger = e^{-\vec{w}^*\cdot\vec{\sigma}/2} \equiv e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2},\;\;\;\;\;\vec{w} = -i\vec{\theta} + \vec{\zeta}
$$
with generators $\vec{J} \rightarrow i\vec{\sigma}/2$, $\vec{K} \rightarrow -\vec{\sigma}/2$ on $(1/2,0)$, and $\vec{J} \rightarrow -i\vec{\sigma}^*/2$, $\vec{K} \rightarrow -\vec{\sigma}^*/2$ on $(0, 1/2)$.
Now look at the sign of the rapidity in the two transformations. It is exactly opposite: if in one rep the transformation is given by a boost $\vec{\zeta}$, in the other rep it is given by the inverse boost $-\vec{\zeta}$. So although both $\Lambda_{(1/2,0)}$ and $\Lambda_{(0,1/2)}$ correspond to the same space-time transform $L$, the change in sign of the boost parameter makes it "as if" they generate opposite formal "rotations": "right-handed" and "left-handed".
Chirality and parity:
On the other hand, the switch from the boost $\vec{\zeta}$ to the inverse $-\vec{\zeta}$ is just the usual relationship between contravariant and covariant transformations. Indeed, for
$$
v'^\mu = L^\mu_{\;\;\nu} v^\nu\;\; \leftrightarrow \;\;v'_\mu = v_\nu(L^{-1})^\nu_{\;\;\mu} = [(L^{-1})^T]_\mu^{\;\;\nu} v_\nu
$$
the exponential forms
$$
L = e^{\vec{\theta}\cdot\vec{J} + \vec{\zeta}\cdot\vec{K}} \;\; \leftrightarrow \;\; (L^{-1})^T = e^{\vec{\theta}\cdot\vec{J} - \vec{\zeta}\cdot\vec{K}}
$$
show that if a contravariant 4-vector transforms under a rotation $\vec{\theta}$ and a boost $\vec{\zeta}$, then the parity transformed covariant 4-vector transforms under the same rotation $\vec{\theta}$ and the inverse boost $-\vec{\zeta}$. Likewise, from the spinor transforms
$$
\Lambda_{(1/2,0)} = e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2}, \;\;\;\;\;\Lambda_{(0,1/2)} = \left(\Lambda_{(1/2,0)}^{-1}\right)^\dagger = e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2}
$$
we see that, in a completely similar way, if a $(1/2,0)$ spinor transforms under a rotation $\vec{\theta}$ and a boost $\vec{\zeta}$, then its $(0, 1/2)$ counterpart transforms under the same rotation $\vec{\theta}$ and the inverse boost $-\vec{\zeta}$. Hence a $(1/2,0)$ (right-handed) spinor is sometimes referred to as a contraspinor, while a $(0,1/2)$ (left-handed) one is then a cospinor (see for instance Andrew Steane's intro to spinors; another nice intro is Schulten's Ch.11 in his QM book).
The contravariant to covariant spinor transformation involves a complex conjugation and is given by
$$
{\hat \chi}\;\;\rightarrow \;\;{\hat \eta} = i\sigma_2 {\hat \chi}^* \equiv \left(\begin{array}{cc}0 & 1 \\-1 & 0\end{array}\right) {\hat \chi}^*
$$
This is so because if ${\hat\chi} \rightarrow \Lambda_{(1/2,0)} {\hat\chi}$ under a Lorentz transformation, then $\hat \eta$ transforms as
$$
{\hat \eta} = i\sigma_2 {\hat \chi}^* \;\;\rightarrow \;\; i\sigma_2 \left(\Lambda_{(1/2,0)} {\hat\chi}\right)^* = i\sigma_2 \Lambda_{(1/2,0)}^* (i\sigma_2)^\dagger (i\sigma_2){\hat\chi}^* = \left(\Lambda_{(1/2,0)}^{-1}\right)^\dagger {\hat \eta} \equiv \Lambda_{(0,1/2)} {\hat \eta}
$$
where the latter equality relies on $(i\sigma_2)^\dagger (i\sigma_2) = I$ and the transformation of the spin matrices $\sigma_j$ under $i\sigma_2$. If we define in addition $\sigma_0 = I$, then for the latter we have
$$
(i\sigma_2) \sigma^*_0(i\sigma_2)^\dagger = \sigma_0
$$
$$
(i\sigma_2)\sigma^*_j (i\sigma_2)^\dagger = - \sigma_j, \;\;\;j=1,2,3
$$
which confirms that $i\sigma_2$ is actually a parity transformation, as expected from the transformation of the $\Lambda$-s,
$$
i\sigma_2 \Lambda_{(1/2,0)}^* (i\sigma_2)^\dagger = \left(\Lambda_{(1/2,0)}^{-1}\right)^\dagger = \Lambda_{(0,1/2)} \;\;\; \leftrightarrow \;\;\;i\sigma_2 \left[e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2}\right]^* (i\sigma_2)^\dagger = e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2}
$$
That is, $i\sigma_2$ leaves the rotation in place, but reverses the direction of the boost - and this is precisely the definition of a space inversion. Which means that $i\sigma_2$ implements not just a change of rep, but a parity transform, just like the metric in Minkowski space (compare the $\Lambda$ relation above with the Lorentz transform relation $gLg = (L^{-1})^T \Leftrightarrow gLg^T = (L^{-1})^T $). Or turning the statement on its ear, applying a parity transform changes the spinor rep, hence its chirality. From this point of view, the right-handed and left-handed terminology can also be understood as labeling spinor transformation properties in right-handed and left-handed 3D reference frames.
But if this is so, why aren't there also 2 chiral irreps for regular 4-vectors? Simply because 4-vectors are real and in this case the contravariant to covariant map is a linear similarity transformation. By Schur's Lemma the reps are then equivalent. Despite the apparent isomorphism with contravariant and covariant representations, respectively, the $(1/2, 0)$ and $(0, 1/2)$ reps are related by an antilinear, not linear, transformation, and formally do not satisfy Schur's Lemma.
Note added in proof: Another simple way to show the connection between chirality and contravariance/covariance, without going into the full blown formalism
Let $\hat \chi$ be any normalized spinor. It is always possible to parametrize it as
$$
{\hat \chi} = \left(\begin{array}{c}\chi^0 \\ \chi^1\end{array}\right) = \left(\begin{array}{c}\chi^0 \\ u/(2\chi^{0*})\end{array}\right), \;\;\; {\hat \chi}^\dagger {\hat \chi} = 1
$$
with $\chi^0, \chi^1\in {\mathbb C}$, $u \equiv 2\chi^{0*}\chi^1 = u^1 + iu^2$, $|u|^2 = 4 |\chi^0|^2(1 - |\chi^0|^2)$ (from $|\chi^0|^2 + |\chi^1|^2 = 1$). The upper contravariant indices are foretelling, but otherwise arbitrary at this point.
The first key observation is that the spin projector associated to $\hat \chi$ reads
$$
{\hat \chi}{\hat \chi}^\dagger = \left(\begin{array}{c}\chi^0 \\ u/(2\chi^{0*})\end{array}\right) \left(\begin{array}{cc}\chi^{0*} & u^*/(2\chi^0)\end{array}\right) = \left(\begin{array}{cc} |\chi^0|^2 & u^*/2 \\ u/2 &1 - |\chi^0|^2) \end{array}\right) = \frac{1}{2} \left(\begin{array}{cc} 1 + \left[\;2|\chi^0|^2-1\;\right] & u^* \\ u &1 - \left[\;2|\chi^0|^2 -1\;\right] \end{array}\right) =
$$
$$
= \frac{1}{2} \left[ I + u^1 \sigma_1 + u^2 \sigma_2 + (2|\chi_0|^2 -1)\sigma_3\right] \equiv \frac{1}{2} \left[ u^0 \sigma_0 + u^1 \sigma_1 + u^2 \sigma_2 + u^3\sigma_3\right]
$$
where $\sigma_\mu$ are Pauli matrices as usual (the lower index has no particular significance), $u^0 = 1$, and $u^3 = 2|\chi_0|^2 -1$, and also
$$
u^\mu = Tr\left(\sigma_\mu {\hat \chi}{\hat \chi}^\dagger\right) = \chi^\dagger \sigma_\mu \chi
$$
A 2nd observation is that we always have
$$(u^1)^2 + (u^2)^2 + (u^3)^2 = |u|^2 + (2|\chi^0|^2 -1 )^2 = 1$$
hence
$$
\det \left({\hat \chi}{\hat \chi}^\dagger \right) = (u^0)^2 - (u^1)^2 - (u^2)^2 - (u^3)^2 = 0
$$
The latter identity obviously suggests that $u^\mu = \chi^\dagger \sigma_\mu \chi$ might behave as a null 4-vector. And indeed, irrespective of the particular rep, under a Lorentz transformation $\Lambda$, $|\det(\Lambda)| = 1$, $\hat \chi$ and its projector transform as
$$
\hat \chi' = \Lambda \hat \chi, \;\;\;{\hat \chi'}({\hat \chi'})^\dagger = \Lambda {\hat \chi}{\hat \chi}^\dagger \Lambda^\dagger
$$
By the same reasoning as above, the transformed projector must again be of the form
$$
{\hat \chi'}({\hat \chi'})^\dagger = \frac{1}{2} \left[ u'^0 I + u'^1 \sigma_1 + u'^2 \sigma_2 + u'^3\sigma_3\right]
$$
$$
u'^\mu = Tr\left[\sigma_\mu {\hat \chi'}({\hat \chi'})^\dagger\right] = (\chi')^\dagger \sigma_\mu \chi'
$$
hence
$$
\det \left[{\hat \chi'}({\hat \chi'})^\dagger \right] = (u'^0)^2 - (u'^1)^2 - (u'^2)^2 - (u'^3)^2 = 0
$$
In other words, $(u^0)^2 - (u^1)^2 - (u^2)^2 - (u^3)^2 = 0$ is a Lorentz invariant, and $u^\mu$ is necessarily a null 4-vector. But it is not clear yet if it is a contravariant or a covariant one. However, from
$$
{\hat \chi'}({\hat \chi'})^\dagger = \Lambda {\hat \chi}{\hat \chi}^\dagger \Lambda^\dagger = \frac{1}{2} \left[ u^0 \Lambda \sigma_0 \Lambda^\dagger + u^1 \Lambda \sigma_1 \Lambda^\dagger + u^2 \Lambda \sigma_2 \Lambda^\dagger + u^3 \Lambda \sigma_3 \Lambda^\dagger \right] \equiv \frac{1}{2} \left[ u'^0 I + u'^1 \sigma_1 + u'^2 \sigma_2 + u'^3\sigma_3\right]
$$
it follows that the transformation law for the $u^\mu$ must be
$$
u'^\mu = \frac{1}{2}\sum_\nu{\left[\text{Tr}\left( \sigma_\mu \Lambda \sigma_\nu \Lambda^\dagger \right) \right] u^\nu}
$$
For the specific forms of $\Lambda_{(1/2,0)}$ and $\Lambda_{(0,1/2)}$ this reads
$$
(1/2, 0):\;\;\;\;\;u'^\mu = \sum_\nu{\left[\text{Tr}\left( \sigma_\mu e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\nu e^{(i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2} \right) \right] u^\nu} =
$$
$$
(0, 1/2):\;\;\;\;\;u'^\mu = \sum_\nu{\left[\text{Tr}\left( \sigma_\mu e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\nu e^{(i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \right) \right] u^\nu}
$$
Since the $u^\mu$ are 4-vectors, the two transformations above must be Lorentz transforms. But notice that if for $(1/2, 0)$ (with no particular placement of the indices)
$$
L_{\mu\nu}(\vec{\theta},\vec{\zeta}) = \text{Tr}\left( \sigma_\mu e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\nu e^{(i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2} \right)
$$
then for $(0, 1/2)$
$$
{\bar L}_{\mu\nu}(\vec{\theta},\vec{\zeta}) = \text{Tr}\left( \sigma_\mu e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\nu e^{(i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \right) = \text{Tr}\left( \sigma_\nu e^{(i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\mu e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \right) = L_{\nu\mu}(-\vec{\theta},-\vec{\zeta}) = \left( \left[ L(\vec{\theta},\vec{\zeta})\right]^{-1}\right)^T_{\mu\nu}
$$
So we always have
$$
(1/2, 0):\;\;\;\;\;u'^\mu = \sum_\nu{L_{\mu\nu}(\vec{\theta},\vec{\zeta}) u^\nu}
$$
$$
(0, 1/2):\;\;\;\;\;u'^\mu = \sum_\nu{ \left( \left[ L(\vec{\theta},\vec{\zeta})\right]^{-1}\right)^T_{\mu\nu} u^\nu}
$$
Compare this with 4-vector contravariant and covariant transforms and it follows that either in $(1/2,0)$ the $u^\mu$ are contravariant and the counterparts in $(0, 1/2)$ are covariant, or conversely. The same applies to any other spinorial bilinears. In other words, we see again that the $(1/2, 0)$ and $(0, 1/2)$ reps are dual to each other. Actually for $(1/2,0)$ it turns out that $(1/2) \text{Tr}\left( \sigma_\mu \Lambda \sigma_\nu \Lambda^\dagger \right) = L^\mu_{\;\;\nu}$, etc, and the corresponding spin 4-vector is indeed contravariant.
A final note on the nature of parity transformed spinors:
The parity transformation $i\sigma_2$, usually referred to as the (antisymmetric) spinor metric $\epsilon \equiv i\sigma_2$, takes a $(1/2, 0)$ spinor ${\hat \chi}$ into its $(0, 1/2)$ equivalent ${\hat \eta} = i\sigma_2 {\hat \chi}^*$. Explicitly this amounts to
$$
{\hat \chi} = \left(\begin{array}{c}\chi^0 \\ u/(2\chi^{0*})\end{array}\right) \rightarrow {\hat \eta} = \left(\begin{array}{cc}0 & 1 \\-1 & 0\end{array}\right)\left(\begin{array}{c}\chi^{0*} \\ u^*/(2\chi^{0})\end{array}\right) = \left(\begin{array}{c}u^*/(2\chi^{0}) \\ -\chi^{0*}\end{array}\right) \equiv \left(\begin{array}{c}{\hat {\bar \chi}}_0\\ -u/2({\hat{\bar\chi}}_0)\end{array}\right)
$$
where ${\hat {\bar \chi}}_0 = u^*/(2{\hat\chi}^0)$. Eventually this shows that the chiral dual ${\hat \eta}$ indeed corresponds to the (space inverted) covariant spin 4-vector $u_\mu = g_{\mu\nu}u^\nu$. But equally important, ${\hat \eta}$ turns out to be the (space inverted) orthogonal of the original ${\hat \chi}$:
$$
{\hat \chi}^\dagger {\hat \eta} = {\hat \chi}^\dagger \epsilon {\hat\chi}^* = \left( {\hat \chi}^T \epsilon {\hat \chi} \right)^* = {\hat \chi}^\dagger \left(\begin{array}{cc}\chi^{0*} & u^*/(2\chi^0)\end{array}\right) \left(\begin{array}{c}u^*/(2\chi^{0}) \\ -\chi^{0*}\end{array}\right) = 0
$$
Bottom line is, chirally dual spinors are the spin-orthogonal, space inverted of each other.
I finally realized that my question was really stupid.
I don't have to change sign from the Poisson super-bracket.
In (pseudo) classical mechanics, given an Lagrangian (second order $L(q,\dot{q})$ in the bosonic case, and first order $L(\xi,\dot{\xi})$ in the fermionic case), the conjugate momentum is given by
$$p=\frac{\partial L}{\partial\dot{q}}$$
$$\pi=L\frac{\overset{\leftarrow}{\partial}}{\partial\dot{\xi}}$$
in the bosonic and fermionic cases, respectively.
The arrow for the fermionic case is due to the anti-commuting nature of Grassmann numbers. Its direction is to make sure that the result should be consistent with the Legendre transformation $$L=\pi\dot{\xi}-H.$$
On the phase space, the Poisson brackets are given by
$$\left\{f(t),g(t)\right\}_{PB}=\frac{\partial f}{\partial q(t)}\frac{\partial g}{\partial p(t)}-\frac{\partial f}{\partial p(t)}\frac{\partial g}{\partial q(t)}$$
$$\left\{\phi(t),\gamma(t)\right\}_{PB}=\phi(t)\left(\frac{\overset{\leftarrow}{\partial}}{\partial\xi(t)}\frac{\overset{\rightarrow}{\partial}}{\partial\pi(t)}+\frac{\overset{\leftarrow}{\partial}}{\partial\pi(t)}\frac{\overset{\rightarrow}{\partial}}{\partial\xi(t)}\right)\gamma(t)$$
for bosonic and fermionic particles, respectively. Since classical mechanics can be viewed as a one-dimensional classical field theory, one can also use functional derivative to re-write them as
$$\left\{f(t),g(t)\right\}_{PB}=\int ds\left\{\frac{\delta f(t)}{\delta q(s)}\frac{\delta g(t)}{\delta p(s)}-\frac{\delta f(t)}{\delta p(s)}\frac{\delta g(t)}{\delta q(s)}\right\}$$
$$\left\{\phi(t),\gamma(t)\right\}_{PB}=\int ds\left[\phi(t)\left(\frac{\overset{\leftarrow}{\delta}}{\delta\xi(s)}\frac{\overset{\rightarrow}{\delta}}{\delta\pi(s)}+\frac{\overset{\leftarrow}{\delta}}{\delta\pi(s)}\frac{\overset{\rightarrow}{\delta}}{\delta\xi(s)}\right)\gamma(t)\right].$$
The above Poisson super-bracket for "classical fermions" is $\mathbb{Z}_{2}$-graded. i.e. $$\left\{\phi,\gamma\right\}_{PB}=-(-1)^{\epsilon(\phi)\epsilon(\gamma)}\left\{\gamma,\phi\right\}_{PB},$$
where where $\epsilon(\,\,\cdot\,\,)$ means the parity of the Grassmann variable ($0$ when even and $1$ when odd).
In field theory, the Poisson super-bracket is generalized to
$$\left\{F(t),G(t)\right\}_{PB}=\int d^{4}x\left\{F(t)\left(\frac{\overset{\leftarrow}{\delta}}{\delta\psi(s,\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\Pi(s,\mathbf{x})}+\frac{\overset{\leftarrow}{\delta}}{\delta\Pi(s,\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\psi(s,\mathbf{x})}\right)G(t)\right\},$$
where $\int d^{4}x\equiv\int ds\int d\mathbf{x}$.
The definition is totally fine. It has a super-commuting property which says $$\left\{F,G\right\}_{PB}=-(-1)^{\epsilon(F)\epsilon(G)}\left\{G,F\right\}_{PB},$$
where $\epsilon(\,\,\cdot\,\,)$ means the parity of the Grassmann variable ($0$ when even and $1$ when odd).
The Lorentzian charge $M_{\mu\nu}$ is clearly Grassmann-even, so their Poisson brackets still anti-commute.
For a Dirac spinor field, the Noether charge of Lorentzian symmetry is given by $$M_{\mu\nu}(t)=-i\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right),$$
where $J_{\mu\nu}(x)=\frac{1}{2}\sigma_{\mu\nu}+i(x_{\mu}\frac{\partial}{\partial x^{\nu}}-x_{\nu}\frac{\partial}{\partial x^{\mu}})𝟙_{4\times 4}$, and the rightarrow means it acts to the right.
Using the above definition of Poisson super-bracket, one has
\begin{align}
&\,\,\,\,\,\,\,\left\{M_{\mu\nu}(t),M_{\alpha\beta}(t)\right\}_{PB} \\
&=-\int d^{4}z\left\{M_{\mu\nu}(t)\left(\frac{\overset{\leftarrow}{\delta}}{\delta\psi(s,\mathbf{z})}\frac{\overset{\rightarrow}{\delta}}{\delta\Pi(s,\mathbf{z})}+\frac{\overset{\leftarrow}{\delta}}{\delta\Pi(s,\mathbf{z})}\frac{\overset{\rightarrow}{\delta}}{\delta\psi(s,\mathbf{z})}\right)M_{\alpha\beta}(t)\right\} \\
&=-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\delta(x-y)\right)\left(\overset{\rightarrow}{J}_{\alpha\beta}(y)\Psi(t,\mathbf{y})\right) \\
&\,\,\,\,\,\,-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right)\Pi(t,\mathbf{y})\left(\overset{\rightarrow}{J}_{\alpha\beta}(y)\delta(x-y)\right) \\
&=-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\delta(x-y)\right)\left(\overset{\rightarrow}{J}_{\alpha\beta}(y)\Psi(t,\mathbf{y})\right) \\
&\,\,\,\,\,\,\,+\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{y})\left(\overset{\rightarrow}{J}_{\alpha\beta}(y)\delta(x-y)\right)\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right).
\end{align}
Since $x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu}$ is anti-symmetric, i.e. its diagonal elements vanishes, one can write it as $\partial^{\nu}x^{\mu}-\partial^{\mu}x^{\nu}$ and integrate by parts. Then, one has, for the orbital part,
\begin{align}
&-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{L}_{\mu\nu}(x)\delta(x-y)\right)\left(\overset{\rightarrow}{L}_{\alpha\beta}(y)\Psi(t,\mathbf{y})\right) \\
&\,\,\,\,\,\,\,+\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{y})\left(\overset{\rightarrow}{L}_{\alpha\beta}(y)\delta(x-y)\right)\left(\overset{\rightarrow}{L}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right) \\
&=\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{x})\left(\overset{\leftarrow}{L}_{\mu\nu}(x)\delta(x-y)\right)\left(\overset{\rightarrow}{L}_{\alpha\beta}(y)\Psi(t,\mathbf{y})\right) \\
&\,\,\,\,\,\,\,-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{y})\left(\overset{\leftarrow}{L}_{\alpha\beta}(y)\delta(x-y)\right)\left(\overset{\rightarrow}{L}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right) \\
&=\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\overset{\leftarrow}{L}_{\mu\nu}(x)\overset{\rightarrow}{L}_{\alpha\beta}(x)\Psi(t,\mathbf{x})-\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\overset{\leftarrow}{L}_{\alpha\beta}(x)\overset{\rightarrow}{L}_{\mu\nu}(x)\Psi(t,\mathbf{x}) \\
&=-\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{L}_{\mu\nu}(x)\overset{\rightarrow}{L}_{\alpha\beta}(x)-\overset{\rightarrow}{L}_{\alpha\beta}(x)\overset{\rightarrow}{L}_{\mu\nu}(x)\right)\Psi(t,\mathbf{x}) \\
&=-\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\left[\overset{\rightarrow}{L}_{\mu\nu},\overset{\rightarrow}{L}_{\alpha\beta}\right]\Psi(t,\mathbf{x}).
\end{align}
Notice that the Lie bracket between the orbital operators is $$[L_{\mu\nu},L_{\alpha\beta}]=i(g_{\sigma\mu}L_{\rho\nu}+g_{\nu\sigma}L_{\mu\rho}-g_{\rho\mu}L_{\sigma\nu}-g_{\nu\rho}L_{\mu\sigma}).$$
Each $L_{\rho\sigma}$ is sandwiched between the conjugate pair of spinor fields and the integral produces a charge $M_{\rho\sigma}$.
For the spin part, the computation is straightforward. Finally, one finds the Lorentz Lie-algebra $$\left\{M_{\mu\nu},M_{\alpha\beta}\right\}_{PB}=i(g_{\sigma\mu}M_{\rho\nu}+g_{\nu\sigma}M_{\mu\rho}-g_{\rho\mu}M_{\sigma\nu}-g_{\nu\rho}M_{\mu\sigma}).$$
Best Answer
Of course these two definitions are not equivalent - one is for a continuum QFT and the other for a lattice theory. A lattice breaks Poincaré invariance, so there's no representations of the Lorentz group to look at for chirality - the notion of chirality doesn't really make sense for a lattice at first glance.
However, the notes by Tong you link in the question are trying to make an argument for why the latter is the correct notion of "chirality on the lattice". Note that Tong is looking at the Hamiltonian of a single fermion. Just prior to eq. (4.28), he discussed the "naive" discretized version of a Weyl fermion in 3+1 dimensions, and shows that the single-particle Hamiltonian is given by $$ H(k)\approx \pm (k-k_0)\cdot \sigma,\tag{1}$$ around points $k_0$ in the Brillouin zone where $\sin(k a) = 0$, where the sign $\pm$ corresponds to the chirality of the continuum fermion prior to discretization.
The generalized fermion on the lattice is really just a straightforward generalization: Instead of the $\sin$, we have arbitrary functions $v_i$, and expanding about a point with $v_i(k_\alpha) = 0$, the first non-vanishing order is $$ H(k) \approx (k-k_\alpha)(\mathrm{D}v(k_\alpha))\sigma,\tag{2}$$ where $\mathrm{D}v$ is the Jacobian matrix of the $v_i$, and by comparing eq. (2) to eq. (1) you see that the sign of $\mathrm{det}(\mathrm{D}v(k_\alpha))$ in eq. (2) corresponds to the $\pm$ in eq. (1), hence is related to the chirality of the fermion.