$\newcommand{\B}{\vec{B}^\times}
\newcommand{\e}{\vec{E}}
\renewcommand{\b}{\vec{\beta}}
\newcommand{\bv}{\vec{B}}$
The field tensor can be written
$\begin{pmatrix}
0 & -\e \\
\e & \B
\end{pmatrix}$,
Where $\B$ is the dual tensor to $\vec{B}$ defined by $\B \vec{v} = \vec{B} \times \vec{v}$. Equivalently, $(\B)_{ik} = \epsilon_{ijk} B_j$. Note $\vec{v}^T \B = (\vec{v} \times \vec{B})^T$. It will also be important to note that $$(\vec{v} \times \vec{w})^\times \vec{u} = \vec{w} (\vec{u} \cdot \vec{v}) - \vec{v} (\vec{u} \cdot \vec{w})$$, so that $$ (\vec{v} \times \vec{w})^\times = \vec{w} \otimes \vec{v} - \vec{v} \otimes \vec{w}$$.
The action of a Lorentz transformation can be written
$$\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}$$, where $$\alpha = \frac{\gamma^2}{1+\gamma}$$. It will be important to note that $$\gamma^2 - \gamma \alpha = \gamma(\gamma - \alpha) = \gamma (\frac{\gamma + \gamma^2 - \gamma^2}{1 + \gamma}) = \frac{\gamma^2}{1+\gamma} = \alpha$$. Also $$1+\alpha \beta^2 = 1 + \alpha (1-1/\gamma^2) =1+ \frac{\gamma^2 -1}{1 + \gamma} =1+ \gamma -1 = \gamma$$
Anyway, the transformed field is
$$\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}
\begin{pmatrix}
0& -\e \\
\e& \B
\end{pmatrix}
\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}$$.
Since the field tensor is antisymmetric, and the Lorentz transformation tensor is symmetric, we know the result must be antisymmetric. We will use this fact later. Let's start by compute the first product
$$\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}
\begin{pmatrix}
0& -\e \\
\e& \B
\end{pmatrix} =
\begin{pmatrix}
-\gamma \b \cdot \e& -\gamma \e-\gamma \b \times \bv \\
\e + \alpha \b (\b \cdot \e)& \gamma \b \otimes \e + \B + \alpha \b \otimes (\b \times \bv)
\end{pmatrix}
$$.
Next we compute the second product. Since we already know this product will be antisymmetric, we will only calculate the right column.
$$
\begin{pmatrix}
-\gamma \b \cdot \e& -\gamma \e-\gamma \b \times \bv \\
\e + \alpha \b (\b \cdot \e)& \gamma \b \otimes \e + \B + \alpha \b \otimes (\b \times \bv)
\end{pmatrix}
\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}
$$
$$
= \begin{pmatrix}
0 & \gamma^2 \b (\b \cdot \e) -\gamma \e-\gamma \b \times \bv -\alpha \gamma \b (\b \cdot \e) \\
\cdots & -\gamma \e \otimes \b - \alpha \gamma (\b \cdot \e) \b \otimes \b + \gamma \b \otimes \e + \B \\
& + \alpha \b \otimes (\b \times \bv) + \alpha \gamma (\b \cdot \e) \b \otimes \b + \alpha (\bv \times \b) \otimes \b
\end{pmatrix}
$$
$$
= \begin{pmatrix}
0 & -(\gamma (\e + \b \times \bv) - (\gamma^2 - \alpha \gamma) \b (\b \cdot \e)) \\
\cdots & \B -\gamma( \e \otimes \b - \b \otimes \e) - \alpha((\b \times \bv) \otimes \b - \b \otimes (\b \times \bv))
\end{pmatrix}
$$
$$
=\begin{pmatrix}
0 & -(\gamma (\e + \b \times \bv) - \alpha \b (\b \cdot \e)) \\
\cdots & \B -\gamma( \b \times \e)^\times - \alpha(\b \times (\b \times \bv))^\times
\end{pmatrix}
$$
By now we have found the expected expression for the new electric field from the upper right entry: $\tilde{\e} =\gamma (\e + \b \times \bv) - \alpha \b (\b \cdot \e)$. Let's now focus on the bottom right entry.
$$ \tilde{\bv}^\times=\B -\gamma( \b \times \e)^\times - \alpha((\b \cdot \bv) \b^\times - \beta^2 \B)$$
$$ = ((1+\alpha \beta^2)\bv - \gamma \b \times \e - \alpha \b (\b \cdot \bv) )^\times$$.
Thus $$\tilde{\bv} = \gamma \bv - \gamma \b \times \e - \alpha \b (\b \cdot \bv) $$
$$ = \gamma(\bv - \b \times \e) - \alpha \b (\b \cdot \bv)$$ as was desired.
In my experience, reading the indices left to right and top to bottom, the first index is the row and the second is the column.
Your screenshot from Carroll doesn't have to be contradictory (although it's definitely confusing/doesn't make rigorous sense). You can just imagine he omits a little "$_{\mu \nu}$" on the matrix:
$$F_{\mu \nu}=\Bigg( \cdots \Bigg)_{\mu \nu}=-F_{\nu \mu}$$
now it's a true real number equation.
Best Answer
Defining $E$ and $B$ in terms of $F$
A little extra generality can help answer question 1, so instead of $1+3$-dimensional spacetime, consider $1+D$-dimensional spacetime, with one time dimension and $D$ spatial dimensions. Then the Faraday tensor $F_{\mu\nu}$ is a $(1+D)\times (1+D)$ antisymmetric matrix. We can think of the Faraday tensor as the primary object, and the electric and magnetic fields are defined in terms of the Faraday tensor like this:
The components $F_{0\mu}=-F_{\mu 0}$ with one time index and one space index are the components of the electric field. This is the definition of the electric field. After accounting for antisymmetry, the electric field has $D$ independent components, as expected for a vector in $D$-dimensional space.
The components with two space indices are the components of the magnetic field. This is the definition of the magnetic field. After accounting for antisymmetry, the magnetic field has $D(D-1)/2$ independent components, which is the number of mutually orthogonal planes in $D$-dimensional space. This is relevant to question 1, but I'll address question 2 first, and then I'll get back to question 1.
Question 2: Why is $F$ antisymmetric?
Question 2 is part of this question: Why is $F_{\mu\nu}$ antisymmetric? There are several ways to help this seem more natural. Here's one: The relativistic version of the Lorentz force equation is $$ \eta_{\mu\nu}\frac{dp^\nu}{d\tau}\propto F_{\mu\nu}p^\nu, \tag{1} $$ where $p^\mu$ is the $1+D$-component energy-momentum vector of a charged particle, $\tau$ is its proper time, and $\eta_{\mu\nu}$ is the Minkowski metric. The quantity $\eta_{\mu\nu}p^\mu p^\nu$ is the invariant mass-squared of the particle, which is independent of $\tau$, so we should have $$ \frac{d}{d\tau} \eta_{\mu\nu}p^\mu p^\nu = 0. \tag{2} $$ By contracting both sides of equation (1) with $p^\mu$, we see that the condition (2) is satisfied if the Faraday tensor is antisymmetric.
Here's another aesthetically appealing consequence of antisymmetry. For a constant-and-uniform electric field, equation (1) is just a steady Lorentz boost of the particle, describing a kind of uniform linear acceleration of the charged particle in the electric field. For a constant-and-uniform magnetic field in a given spatial plane, equation (1) is just a steady rotation of the particle's momentum, describing constant-speed circular motion in that same spatial plane. The antisymmetry of $F_{\mu\nu}$ is responsible for these simple behaviors.
Here's yet another way to appreciate the antisymmetry. If we didn't impose any symmetry requirements on $F_{\mu\nu}$, then we could decompose it into separate parts that don't mix with each other under Lorentz transformations. By that criterion, we should treat those parts as different kinds of fields. In particular, the antisymmetric part, the traceless symmetric part, and the remainder (a scalar) would all qualify as different types of fields. One of these, the antisymmetric part, has just the right properties to comply with everything we know about the electromagnetic field, such as the properties highlighted above.
Question 1: Why are the components of $B$ arranged that way in $F$?
Why are the $3$ components of the magnetic field all jumbled up (signs and positions) in the $D=3$ version of the Faraday tensor? This is where the generalization to arbitrary $D$ is enlightening. For arbitrary $D$, we see that the magnetic field has $D(D-1)/2$ independent components, which is not the right number of components for a spatial vector! That's becasue the magnetic field is not a spatial vector: it's a bivector, which can be visualized as a directed element of area rather than a directed element of length. In the physically relevant case $D=3$, the two are interchangeable, because we there is a unique line orthogonal to any given plane. That's a special feature of $3$-dimensional space that doesn't generalize to higher (or lower) dimensions. What we should do is write the magnetic field with two indices: $B_{jk}=-B_{kj}$, where the two indices match the two axes of the corresponding plane. But drawing directed elements of area is difficult, so in $D=3$, we use a trick: we define $$ B_1 := B_{23} \hskip2cm B_2 := B_{31} \hskip2cm B_3 := B_{12} \tag{3} $$ and treat $B_k$ as though they were the components of a vector instead of a bivector. Notice the cyclic pattern in the indices: that turns out to correspond to the idea of a line orthogonal to the given plane. One of the things that makes this trick work is that in $D=3$, vectors and bivectors transform the same way under rotations, via the correspondence (3). However, we see a difference when we consider reflections: a vector changes sign when we reflect all three dimensions of space, but a bivector doesn't. The magnetic field is a bivector, and it does not change sign when we reflect all three dimensions of space. Sometimes people call it an axial vector or a pseudovector, which in general is the name for a completely antisymmetric quantity with $D-1$ indices. For $D=3$, that's the same as a bivector.
The correspondence (3), together with the antisymmetry $B_{jk}=-B_{kj}$, explains the pattern of positions and signs for the components of the magnetic field in the $D=3$ version of the Faraday tensor.