Why is a Lorentz transformation of a Lorentz transformation also a Lorentz transformation? Why is a Lorentz transformation of a Lorentz transformation ($x''$,$y''$,$z''$,$t''$) also a Lorentz transformation?
 A: I'll assume that the motion is in the only x-direction.
You can define the Lorentz transformation as 
$$P^{\overline{\alpha}} = \Lambda^{\overline{\alpha}}_\beta~P^{\beta}~~(1)$$
where $P^{\overline{\alpha}} =(P^{\overline{0}}P^{\overline{1}},P^{\overline{2}},P^{\overline{3}})$(represents the coordinate of an event measured by $O'$, which moves with a velocity $v_1$ with respect to an observer $O$) and $P^{\beta} = (P^0, P^1, P^2, P^3)$ (represents the coordinate of an event measured by $O$).
$$\Lambda^{\overline{\alpha}}_\beta = \begin{pmatrix}
\gamma_1 & -v_1\gamma_1 & 0 & 0 \\
-v_1\gamma_1 & \gamma_1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}$$
Let us write a Lorentz transformation between $O''$ and $O'$. 
$$P^{\overline{\overline{\nu}}} = \Lambda^{\overline{\overline{\nu}}}_{\overline{\alpha}}~ P^{\overline{\alpha}}~~(2)$$
Here $P^{\overline{\overline{\nu}}}$ represents the coordinate of an event measured by $O''$, which moves with a velocity $+v_2$ to an observer $O'$.
In this case 
$$\Lambda^{\overline{\overline{\nu}}}_{\overline{\alpha}}  = \begin{pmatrix}
\gamma_2 & -v_2\gamma_2 & 0 & 0 \\
-v_2\gamma_2 & \gamma_2 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}$$
We can combine Eqn $(1)$ and $(2)$ to write
$$P^{\overline{\overline{\nu}}} = \Lambda^{\overline{\overline{\nu}}}_{\overline{\alpha}} ~\Lambda^{\overline{\alpha}}_\beta~P^{\beta}$$
Where $$\Lambda^{\overline{\overline{\nu}}}_{\overline{\alpha}} ~\Lambda^{\overline{\alpha}}_\beta =\Lambda^{\overline{\overline{\nu}}}_\beta =\begin{pmatrix}
\gamma_2 & -v_2\gamma_2 & 0 & 0 \\
-v_2\gamma_2 & \gamma_2 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}\begin{pmatrix}
\gamma_1 & -v_1\gamma_1 & 0 & 0 \\
-v_1\gamma_1 & \gamma_1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}$$
As you can see $\Lambda^{\overline{\overline{\nu}}}_\beta$ is also a Lorentz transformation matrix, which is the multiplication of the two transformation matrices. 
Thus we can write
$$P^{\overline{\overline{\nu}}} = \Lambda^{\overline{\overline{\nu}}}_\beta P^{\beta}$$
In principle, you can extend this transformation for any observer which is an inertial reference frame.
A: You're asking the wrong question. You don't apply a Lorentz transformation (LT) to a transformation, you apply a Lorentz transformation to an inertial reference frame (1) to get a description of an event (or events) in another inertial reference frame (2). That LT will preserve the space-time quantity $$c^2t_1^2-r_1^2 = c^2t_2^2-r_2^2.$$
If that is true (and it is) it doesn't matter what the initial frame is. You can always apply LT to get a different frame that preserves. Each LT will "boost" a frame (non-linearly related to $\beta$ -- $\beta=\tanh s$). And the application of an LT twice or thrice or more always produces the equivalent of a single LT with a matching boost sum.
Edit Addition: 
If the relative velocity between two frames is $\beta c$, then we can represent \begin{align}
\gamma &= \cosh s \newline
\gamma\beta &= \sinh s \end{align}
where $s$ is called the boost of the LT.
It can be shown (a good exercise for those in COVID-19 isolation) that $$LT(s_1)LT(s_2) = LT(s_1+s_2).$$
A: According to Landau&Livshitz Lorentz transformation is a (hyperbolic) rotation in time-space. Rotation after a rotation is also a rotation.
A: All the Lorentz transformations of Minkowski spacetime form a group -- Lorentz group. By closure law, LT followed by LT is just another LT.
A: I would say that you want a mathematical proof. You can see in Reign's answer a particular case, but here I will post a more general one.
As you may know, a Lorentz transformation can be written as a 4-dimensional matrix, $\Lambda$, so that $x' = \Lambda x$. This matrix must fulfill this relation,
\begin{equation}
\Lambda^T \eta \Lambda = \eta
\end{equation}
where $\eta$ is the Minkowski metric.
If you consider a Lorentz transformation you are thinking of $x' = \Lambda_1 x$. A Lorentz transformation of a Lorentz transformation is then $x'' = \Lambda_2 x' = \Lambda_2 \Lambda_1 x$. Now, you have a transformation matrix $\Lambda = \Lambda_2 \Lambda_1$ and you would like to know whether this new matrix $\Lambda$ is a Lorentz transformation or not. To do this, take the previous equation and do some calculations:
\begin{equation}
\Lambda^T \eta \Lambda = (\Lambda_2 \Lambda_1)^T \eta (\Lambda_2 \Lambda_1) = (\Lambda_1)^T (\Lambda_2)^T \eta \Lambda_2 \Lambda_1
\end{equation}
Now, you can use that $(\Lambda_2)^T \eta \Lambda_2 = \eta$ because $\Lambda_2$ is a Lorentz transformation, so that you get
\begin{equation}
\Lambda^T \eta \Lambda = (\Lambda_1)^T \eta \Lambda_1
\end{equation}
For the same reason you can use that $(\Lambda_1)^T \eta \Lambda_1 = \eta$. 
Finally you get 
\begin{equation}
\Lambda^T \eta \Lambda = \eta
\end{equation}
where $\Lambda = \Lambda_2 \Lambda_1$. This proofs that a Lorentz transformation of a Lorentz transformation is a Lorentz transformation.
