= Solution
Use <natural units> and the <Minkowski metric> $g=\operatorname{diag}(1,-1,-1,-1)$. The <Lorentz group> is
$$
G=O(1,3)=\{L\in GL(4,\mathbb R):L^TgL=g\}.
$$
Write a transformation near the identity as $L=I+\Omega+O(\Omega^2)$. The defining relation gives $\Omega^Tg+g\Omega=0$, so $\omega_{\mu\nu}=g_{\mu\rho}\Omega^\rho{}_\nu$ is antisymmetric. Choose
$$
(M^{\rho\sigma})^\mu{}_\nu
=g^{\rho\mu}\delta^\sigma_\nu-g^{\sigma\mu}\delta^\rho_\nu,
\qquad
\Omega=\frac12\omega_{\rho\sigma}M^{\rho\sigma}.
$$
There are $\binom42=\boxed{6}$ independent real parameters: three spatial rotations and three <Lorentz boosts>. Differentiating $(e^{t\Omega})^Tg e^{t\Omega}$ with respect to $t$ shows it is constant and equals $g$. Thus exponentiating these infinitesimal generators gives <Lorentz transformations> connected to the identity. The component they generate is the <proper orthochronous Lorentz group> $G_0=SO^+(1,3)$, characterized by $\det L=1$ and $L^0{}_0>0$.
The local exponential calculation alone establishes products of exponentials. The single-exponential assertion also holds here. Use the <Lorentz spinor double cover> $SL(2,\mathbb C)\to SO^+(1,3)$, obtained by acting on $X=x^0I+x^a\sigma_a$ as $X\mapsto AXA^\dagger$. Its <determinant> is the Minkowski norm. A lift $A$ with distinct eigenvalues $\lambda,\lambda^{-1}$ is diagonalizable and has a traceless logarithm with eigenvalues $z,-z$, where $e^z=\lambda$. With repeated eigenvalues, choose between $A$ and $-A$ so that the lift is $I+N$ with $N^2=0$; its traceless logarithm is $N$. Since the two signs have the same Lorentz image, exponentiating the differential of the covering map proves <every proper orthochronous Lorentz transformation is an exponential>:
$$
\boxed{L=\exp\left(\frac12\omega_{\rho\sigma}M^{\rho\sigma}\right).}
$$
It is not a global assertion about the disconnected improper components of $G$.
For <Dirac spinors>, choose <gamma matrices> satisfying the <Clifford algebra> $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I$ and set
$$
\Sigma^{\rho\sigma}=\frac14[\gamma^\rho,\gamma^\sigma],
\qquad S=\exp\left(\frac12\omega_{\rho\sigma}\Sigma^{\rho\sigma}\right).
$$
The Clifford relations give
$$
[\Sigma^{\rho\sigma},\gamma^\mu]
=g^{\sigma\mu}\gamma^\rho-g^{\rho\mu}\gamma^\sigma.
$$
They also give the stipulated Lorentz <commutator> with every $M$ replaced by $\Sigma$: commute $\Sigma$ through the two gamma factors defining the other generator and collect the four terms. Consequently this is a representation of the <Lie algebra of the Lorentz group>, and exponentiation yields
$$
S^{-1}\gamma^\mu S=L^\mu{}_\nu\gamma^\nu,\qquad
\psi'(x')=S\psi(x),\quad x'=Lx.
$$
The <Dirac adjoint> $\bar\psi=\psi^\dagger\gamma^0$ transforms as $\bar\psi'=\bar\psi S^{-1}$, since $S^\dagger\gamma^0S=\gamma^0$.
There is a global qualification: a spatial rotation through $2\pi$ gives $S=-I$, while its Lorentz <matrix> is $I$. Thus \b[<Dirac spinors> form an ordinary representation of the double cover, and a double-valued representation of $G_0$], rather than a single-valued linear representation of $G_0$ itself. <Fermion bilinears> are unaffected by this sign.
For a <scalar field>, $\phi'(x')=\phi(x)$ and its derivative transforms as a covector. A <Lagrangian density> must have all Lorentz indices contracted into scalars using invariant tensors; arbitrary scalar potentials are allowed. For example the kinetic contraction and a potential $V(\phi)$ are invariant. Lorentz invariance by itself does not restrict the potential to quartic order: that restriction would be an additional four-dimensional power-counting requirement.
Define $\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3$. It commutes with all $\Sigma^{\rho\sigma}$, so the <axial current> $j_5^\mu=\bar\psi\gamma^\mu\gamma^5\psi$ transforms as a vector under $G_0$. Hence \b[all three proposed contractions are invariant under $G_0$]. If invariance under parity or the full improper <Lorentz group> is also required, the distinction is:
$$
\begin{array}{c|c|c}
\text{operator}&G_0&\text{parity, for ordinary scalar }\phi\\ \hline
\partial_\mu\phi\,\partial^\mu\phi&\text{invariant}&\text{even}\\
\partial_\mu\phi\,j_5^\mu&\text{invariant}&\text{odd}\\
j_{5\mu}j_5^\mu&\text{invariant}&\text{even}.
\end{array}
$$
The <axial current> is a <pseudovector>: under an improper <Lorentz transformation> it has the extra <determinant> sign. The mixed derivative–axial contraction is therefore a <pseudoscalar>, while the square of the <axial current> has two cancelling signs. If $\phi$ is itself a <pseudoscalar>, its derivative supplies the second sign and the mixed coupling becomes parity even. This is the <axial derivative coupling and parity> distinction; parity invariance must not be silently assumed from proper Lorentz invariance.
In four dimensions, $[\phi]=1$, $[\psi]=3/2$ and $[\partial_\mu]=1$. The three operators have <mass dimensions> $4,5,6$, respectively. Thus the latter two require dimensionful couplings and are not power-counting renormalizable interactions, but this does not invalidate their Lorentz invariance or their use in an <effective field theory>.
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