The short answer to your first question is that the Newman-Penrose Maxwell equations are the different components of the general relativistic Maxwell equations in spinor form.
To see this, note that your second equation ($F_{[ab;c]} = 0$) can be expressed as ${}^*F^{ab}{}_{;b} = 0$, similar to the first one $F^{ab}{}_{;b} = 0$ (these are of course the source free equations). Then, using the symmetries of the field tensor ($F_{ab} = F_{[ab]}$) we can expand it in spinor form as (see below for a derivation if you are unfamiliar with spinor algebra):
$$
F_{ab} = \phi_{AB}\epsilon_{A'B'} + \overline{\phi}_{A'B'}\epsilon_{AB},
$$
where $\phi_{AB} = \phi_{(AB)}$, and $\epsilon_{AB}$ is the spinor metric (Levi-Civita symbol). And since the dual in spinor form becomes (again see further below for a derivation)
$$
{}^*F_{ab} = i(\phi_{AB}\epsilon_{A'B'} - \overline{\phi}_{A'B'}\epsilon_{AB}),
$$
the two Maxwell equations in spinor form reduce to
$$\tag{1}
\phi^{AB}{}_{;A'B} = 0.
$$
Now, since the field spinor $\phi_{AB}$ is symmetric its independent components are given by
$$
\begin{aligned}
\phi_0 &\equiv \phi_{AB}o^Ao^B, &
\phi_1 &\equiv \phi_{AB}o^A\iota^B, &
\phi_2 &\equiv \phi_{AB}\iota^A\iota^B, \\
&= F_{ab}l^am^b, &
&= \frac{1}{2}F_{ab}(l^an^b + \overline{m}^am^b), &
&= F_{ab}\overline{m}^an^b, \\
&= F_{13}, &
&= \frac{1}{2}(F_{12} + F_{43}), &
&= F_{42},
\end{aligned}
$$
where $(o^A,\iota^A)$ is our spinor dyad. Here the second row follows by the identification of null tetrad with spinor pairs $(l^a,n^a,m^a,\overline{m}^a) = (o^Ao^{A'},\iota^A\iota^{A'},o^A\iota^{A'},\iota^Ao^{A'})$, and the third line is obtained by simply noting that e.g. $l^a$ is the first null frame basis vector and $m^a$ is the third so $F_{ab}l^am^b = F_{13}$ by definition.
The components of the spinor Maxwell equation $(1)$ will include directional derivatives of the components along with products of the components and different spin-coefficients, according to the standard definition of the covariant derivative (which Chandrasekhar calls intrinsic derivatives). The scalar versions are given on the next page in his book, so I will not write them out here.
To clarify then: the four spinor equations you give form a system of equations equivalent to the source-free Maxwell equations. So you are on the right track in your last question, but the equations are in fact formed from both $F^{ab}{}_{;b} = 0$ and $F_{[ab;c]} = 0$.
Expanding the field tensor in spinor form: Using the fact that the spinor metric $\epsilon_{AB}$ is skew-symmetric (it is usually taken to be normalized as the Levi-Civita symbol, but this is not necessary) we immediately get
$$
\epsilon_{AB}\epsilon_{CD} + \epsilon_{AD}\epsilon_{BC} + \epsilon_{AC}\epsilon_{DB} = 0,
$$
or more suggestively, upon raising a few indices
$$
\delta_A^C\delta_B^D - \delta_B^C\delta_A^D = \epsilon_{AB}\epsilon^{CD}.
$$
So for some spinor $Q_{\cdots AB \cdots}$ we get by spliting the index pair $AB$ into the symmetric and skew-symmetric part
$$\tag{2}
Q_{\cdots AB \cdots} = Q_{\cdots (AB) \cdots} + \frac{1}{2}\epsilon_{AB}Q_{\cdots C}{}^C{}_{\cdots}.
$$
Doing some explicit algebra on the spinor form of an skew-symmetric tensor $T_{ab} = T_{[ab]}$ we get
\begin{align}
T_{ab} &= \frac{1}{2}(T_{ab} - T_{ba}) \\
&= \frac{1}{2}(T_{AA'BB'} - T_{BB'AA'}) \\
&= \frac{1}{2}\left((T_{ABA'B'} - T_{ABB'A'}) + (T_{ABB'A'} - T_{BAB'A'})\right) \tag{3}\\
&= T_{(AB)[A'B]} + T_{[AB](A'B')} \\
&= \frac{1}{2}(T_{(AB)C'}{}^{C'}\epsilon_{A'B'} + T_{(A'B')C}{}^C\epsilon_{AB}),
\end{align}
where the symmetrization of the remaining index pair follows since the trace of an anti-symmetric tensor is zero. If $T_{ab}$ is real, the two terms must be complex conjugates. So for $F_{ab}$ we define $\phi_{AB} = \frac{1}{2}F_{(AB)C'}{}^C$. To get the desired form.
The spinor form of the dual: To do this we must note that the spinor equivalent of a tensor is given by maps $\varsigma_p : T_pM \to (\mathcal{S} \otimes \overline{\mathcal{S}})$, where $\mathcal{S}$ is our spin space. Dropping, the $p$ subscript and adopting index notation for this map, we demand that $\varsigma$ satisfies
$$\tag{4}
g_{ab} = \varsigma_a{}^{AA'}\varsigma_b{}^{BB'}\epsilon_{AB}\epsilon_{A'B'}.
$$
The dual ${}^*T_{ab}$ of a skew-symmetric tensor $T_{ab} = T_{[ab]}$ is defined as
\begin{align*}
{}^*T_{ab} = \frac{1}{2}\sqrt{-g}\epsilon_{abcd}T^{cd},
\end{align*}
where $\epsilon_{abcd}$ is the Levi-Civita symbol and $g$ is the determinant of the metric. To write this in spinor form note that the product $\sqrt{-g}\epsilon_{abcd}$ forms the Levi-Civita (pseudo-)tensor, whence from the definition
\begin{align}\tag{5}
\sqrt{-g}\epsilon_{abcd} &= \sqrt{-g}\epsilon_{abcd}\varsigma^a{}_{AA'}\varsigma^b{}_{BB'}\varsigma^c{}_{CC'}\varsigma^d{}_{DD'} \\
&= \sqrt{-g}\varsigma\epsilon_{AA'BB'CC'DD'},
\end{align}
where $\varsigma$ is the determinant of $\varsigma^a{}_{AA'}$ with $AA'$ treated as a single vector index, and $\epsilon_{AA'BB'CC'DD'}$ correspondingly is the Levi-Civita symbol with the indices treated the same way. Now, from the skew-symmetry of $\epsilon_{AB}$ we have $\epsilon_{AB} = \lambda\widetilde{\epsilon}_{AB}$ for some $\lambda$, where $\widetilde{\epsilon}_{AB}$ is the Levi-Civita symbol. Therefore, one can easily verify that
\begin{align}\tag{6}
\pm|\lambda|^4\epsilon_{AA'BB'CC'DD'} &= \epsilon_{AD}\epsilon_{BC}\epsilon_{A'C'}\epsilon_{B'D'} - \epsilon_{AC}\epsilon_{BD}\epsilon_{A'D'}\epsilon_{B'C'},
\end{align}
with the sign dependent on how the orientation of $TM$ is related to the spinor index pairs by $\varsigma_a{}^{AA'}$. We can always choose to have e.g. a plus sign (which corresponds to the construction $(l^a,n^a,m^a,\overline{m}^a) = (o^Ao^{A'},\iota^A\iota^{A'},o^A\iota^{A'},\iota^Ao^{A'})$ or its equivalents), but in either case $(4)$ and $(6)$ yield $g = \varsigma^{-2}|\lambda|^8$. Plugging this result into $(5)$, using $(6)$ again, and raising $c$ and $d$ finally yields
\begin{align}\tag{7}
\sqrt{-g}\epsilon_{abef}g^{ec}g^{df} = \pm i\left(\delta^D_A\delta^C_B\delta^{C'}_{A'}\delta^{D'}_{B'} - \delta^C_A\delta^D_B\delta^{D'}_{A'}\delta^{C'}_{B'}\right),
\end{align}
whence we can write
\begin{align*}
{}^*T_{ab} &= \pm\frac{i}{2}\left(T_{(AB)C'}{}^{C'}\epsilon_{A'B'} - {T}_{(A'B')C}{}^C\epsilon_{AB}\right),
\end{align*}
by applying $(7)$ to $(3)$, in effect adding $\pm i$ as a factor to $\epsilon_{A'B'}$ and $\mp i$ as a factor to $\epsilon_{AB}$. Using the previously derived definition of $\phi_{AB}$, and selecting a "$+$-orientation," we get the desired expression.
