Definition of Transpose is $(A^T)_{ij} = A_{ji}$
$1.$ Why $\begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M^T \\ N^T \end{bmatrix}$, and NOT $\begin{bmatrix} M \\ N\end{bmatrix}$?
After the transpose, $M$ is in (1, 1) position and $N$ is in (2,1) position. Why still keep the $^T$?
For example, pursuant to Git Gud's comment, I tried $\begin{bmatrix}
\begin{bmatrix}
1 & 5 \\
3 & 7 \\
\end{bmatrix}
&
\begin{bmatrix}
2 & 6 \\
4 & 8 \\
\end{bmatrix}
\end{bmatrix}^T $.
This means transposing the entries (the two 2 by 2 matrices) once, so: $ \begin{bmatrix} \begin{bmatrix}
1 & 5 \\
3 & 7 \\
\end{bmatrix} \\
\begin{bmatrix}
2 & 6 \\
4 & 8 \\
\end{bmatrix} \end{bmatrix}$ ?
$2.$ Why $\begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M^T & N^T \end{bmatrix}$, and NOT $ \begin{bmatrix} M & N \end{bmatrix}$ ?
After transpose - $M$ is in (1, 1) position and $N$ is in (1,2) position. Why still keep the $^T$?
Per contra, this example contains not one $^T$ at the end. I recast it as a question.
I see user Eike Schulte's comment and brook that $c^T = c$ for all complex numbers $c$.
$\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}^T = \begin{bmatrix} a_{11}^T & a_{12}^T \\ a_{21}^T & a_{22}^T \end{bmatrix} = \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix}$
Given two matrices $M_{m\times n},N_{m\times p}$, there are two ways to interpret the entity $\begin{bmatrix} M & N \end{bmatrix}$.
One is the $m\times (n+p)$ matrix whose $(i,j)$ entry is $\begin{cases} (M)_{(i,j)}, &\text{if }j\leq n\\ (N)_{(i, j-n)}, &\text{if }j\ge n+1\end{cases}$.
In this case I'd rather denote the matrix described above as $\begin{bmatrix} M \mid N \end{bmatrix}$, (the augmented matrix). This is standard notation.
The other is a $1\times 2$ matrix whose first entry is the matrix $M$ and whose second entry is the matrix $N$.
Under the first interpretation one has $$\begin{bmatrix} M & N \end{bmatrix}^T=\begin{bmatrix} M \mid N \end{bmatrix}^T=\begin{bmatrix} M^T \\ \overline{N^T} \end{bmatrix}.$$
Under the second interpretation one has $$\begin{bmatrix} M & N \end{bmatrix}^T=\begin{bmatrix} M \\ N \end{bmatrix}.$$
The second interpretation is very uncommon. Most of the time it's safe to assume one is under the first interpretation.
An example: Let $M=\begin{bmatrix} 1 & 0 & 1\\ 2 & 3 & 5\end{bmatrix}_{2\times 3}$ and $N=\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}_{2\times 2}$.
The first interpretation yields the matrix $A$ where $$(A)_{ij}=\begin{cases} (M)_{(i,j)}, &\text{if }j\leq 3\\ (N)_{(i, j-n)}, &\text{if }j\ge 4\end{cases}, \text{ for all }(i,j)\in \{1,2\}\times\{1,2,3,4,5\}.$$
That is $$ A=\left[\begin{array}{ccc|cc} (M)_{11} & (M)_{12} & (M)_{13} & (N)_{14} & (N)_{15}\\ (M)_{21} & (M)_{22} & (M)_{23} & (N)_{24} & (N)_{25} \end{array}\right]=\left[\begin{array}{ccc|cc} 1 & 0 & 1 & 1 & 1\\ 2 & 3 & 5 & 0 & 1 \end{array}\right].$$
Transposing yields $A^T=\left[\begin{array}{cc}1 & 2\\ 0 & 3\\ 1 & 5\\ \hline 1 & 0\\ 1 & 1 \end{array}\right]=\left[\begin{array}{c}M^T\\ \hline N^T \end{array}\right]$.
The second interpretation gives $A_{1\times 2}=\left[\begin{matrix} (A)_{11} & (A)_{12}\end{matrix}\right]_{1\times 2}$ where $(A)_{11}=M$ and $(A)_{12}=N$, transposing: $$\left(A^T\right)_{2\times 1}=\begin{bmatrix} (A)_{11}\\ (A)_{21}\end{bmatrix}_{2\times 1}=\begin{bmatrix} M\\ N\end{bmatrix}_{2\times 1}=\begin{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ 2 & 3 & 5\end{bmatrix}_{2\times 3}\\ \begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}_{2\times 2} \end{bmatrix}_{2\times 1}.$$
Here the entries in the matrix just happen to be matrices themselves, you can create matrices in which their entries are whatever you want. For instance, $\begin{bmatrix} 1 & \begin{bmatrix} 1 & 2\\ 3 & 4\end{bmatrix} & \spadesuit\\ \implies & \huge{〠} & +\end{bmatrix}$ is a matrix.