$$
(1) :\begin{cases}
\boldsymbol{\dot x}_1=\boldsymbol{A}_1 \boldsymbol{x}_1+\boldsymbol{B}_1 \boldsymbol{u}_1 \\
\boldsymbol{y}_1=\boldsymbol{C}_1 \boldsymbol{x}_1+\boldsymbol{D}_1 \boldsymbol{u}_1
\end{cases}
$$
$$
(2):
\begin{cases}
\boldsymbol{\dot x}_2=\boldsymbol{A}_2 \boldsymbol{x}_2+\boldsymbol{B}_2 \boldsymbol{u}_2 \\
\boldsymbol{y}_2=\boldsymbol{C}_2 \boldsymbol{x}_2+\boldsymbol{D}_2 \boldsymbol{u}_2
\end{cases}
$$
I assume you mean the output of the first system is the input of the second system which means
$$
\boldsymbol{u}_2=\boldsymbol{y}_1
$$
Then you can reform the second system as
$$
(2b)\begin{cases}
\boldsymbol{\dot x}_2=\boldsymbol{A}_2 \boldsymbol{x}_2+\boldsymbol{B}_2 (\boldsymbol{C}_1 \boldsymbol{x}_1+\boldsymbol{D}_1 \boldsymbol{u}_1) \\
\boldsymbol{y}_2=\boldsymbol{C}_2 \boldsymbol{x}_2+\boldsymbol{D}_2 (\boldsymbol{C}_1 \boldsymbol{x}_1+\boldsymbol{D}_1 \boldsymbol{u}_1)
\end{cases}
$$
Combining (1) and (2b):
$$
\begin{cases}
\boldsymbol{\dot x}_1=\boldsymbol{A}_1 \boldsymbol{x}_1+\boldsymbol{B}_1 \boldsymbol{u}_1 \\
\boldsymbol{\dot x}_2=\boldsymbol{A}_2 \boldsymbol{x}_2+\boldsymbol{B}_2 \boldsymbol{C}_1\boldsymbol{x}_1+\boldsymbol{B}_2 \boldsymbol{D}_1 \boldsymbol{u}_1 \\
\boldsymbol{y}_2=\boldsymbol{C}_2 \boldsymbol{x}_2+\boldsymbol{D}_2 \boldsymbol{C}_1 \boldsymbol{x}_1+\boldsymbol{D}_2\boldsymbol{D}_1 \boldsymbol{u}_1
\end{cases}
$$
Let's make it neat:
$$\begin{cases}
&\begin{bmatrix}\boldsymbol{\dot x}_1\\ \boldsymbol{\dot x}_2\end{bmatrix}&=\begin{bmatrix}
\boldsymbol{A}_1 & \boldsymbol{0}\\
\boldsymbol{B}_2\boldsymbol{C}_1 & \boldsymbol{A}_2
\end{bmatrix}\begin{bmatrix}\boldsymbol{x}_1\\\boldsymbol{x}_2\end{bmatrix}
+
\begin{bmatrix}
\boldsymbol{B}_1\\
\boldsymbol{B}_2\boldsymbol{D}_1
\end{bmatrix}\boldsymbol{u}_1 \\
&\boldsymbol{y}_2&=\begin{bmatrix}\boldsymbol{D}_2\boldsymbol{C}_1&\boldsymbol{C}_2\end{bmatrix}\begin{bmatrix}\boldsymbol{x}_1\\\boldsymbol{x}_2\end{bmatrix}+\begin{bmatrix}\boldsymbol{D}_2\boldsymbol{D}_1\end{bmatrix}\boldsymbol{u}_1
\end{cases}
$$
Compacting as a new system:
$$
\begin{cases}
\boldsymbol{\dot x}=\boldsymbol{A} \boldsymbol{x}+\boldsymbol{B} \boldsymbol{u} \\
\boldsymbol{y}=\boldsymbol{C} \boldsymbol{x}+\boldsymbol{D} \boldsymbol{u}
\end{cases}
$$
$$
\begin{align}
& \boldsymbol{A}=\begin{bmatrix}
\boldsymbol{A}_1 & \boldsymbol{0}\\
\boldsymbol{B}_2\boldsymbol{C}_1 & \boldsymbol{A}_2
\end{bmatrix}\\
& \boldsymbol{B}=\begin{bmatrix}
\boldsymbol{B}_1\\
\boldsymbol{B}_2\boldsymbol{D}_1
\end{bmatrix}\\
& \boldsymbol{C}=\begin{bmatrix}\boldsymbol{D}_2\boldsymbol{C}_1&\boldsymbol{C}_2\end{bmatrix} \\
& \boldsymbol{D}=\begin{bmatrix}\boldsymbol{D}_2\boldsymbol{D}_1\end{bmatrix}
\end{align}
$$
The expression $\hat G(s) = [\hat G_1; \hat G_2]$ is a transfer function matrix, which is generalization of SISO transfer function to MIMO. For example when both $\hat G_1$ and $\hat G_2$ are SISO, such as
$$
\hat G_1(s) = \frac{1}{s}, \quad \hat G_2(s) = \frac{1}{s+1}
$$
you get
$$
\hat G(s) =
\begin{bmatrix}
\frac{1}{s} \\ \frac{1}{s+1}
\end{bmatrix}
$$
which represents a system with one input and two outputs.
You can also stack state space models which is equivalent to $\hat G(s) = [\hat G_1; \hat G_2]$ by defining the state vector as $x=\begin{bmatrix}x_1^\top & x_2^\top\end{bmatrix}^\top\!\!\in\mathbb{R}^{n_1+n_2}$ with the following state space model
$$
A = \begin{bmatrix}A_1 & 0 \\ 0 & A_2\end{bmatrix}, \quad
B = \begin{bmatrix}B_1 \\ B_2\end{bmatrix}, \quad
C = \begin{bmatrix}C_1 & 0 \\ 0 & C_2\end{bmatrix}, \quad
D = \begin{bmatrix}D_1 \\ D_2\end{bmatrix}.
$$
If instead you want $\hat G(s) = [\hat G_1, \hat G_2]$ (so for SISO $\hat G_1$ and $\hat G_2$ you would have that $\hat G$ has two inputs and one output) you can use
$$
A = \begin{bmatrix}A_1 & 0 \\ 0 & A_2\end{bmatrix}, \quad
B = \begin{bmatrix}B_1 & 0 \\ 0 & B_2\end{bmatrix}, \quad
C = \begin{bmatrix}C_1 & C_2\end{bmatrix}, \quad
D = \begin{bmatrix}D_1 & D_2\end{bmatrix}.
$$
However this method of combining state space models might not yield minimal models, so they might either be uncontrollable or unobservable. This will for example be the case when $\hat G_1 = \hat G_2$.
Best Answer
To realize an improper transfer function, derivatives of the input would be needed. The answer above by Rodrigo de Azevedo helps make clear why. The problem is that it is not possible to realize perfect derivatives. A number of arguments are helpful in understanding why.
The modulus of the frequency response of a differentiator increases with frequency. However it is not possible to construct an apparatus whose gain becomes arbitrary large at large frequencies. On the contrary, any device known will have a cutoff frequency after which its response falls.
Or, suppose you feed a discontinuous signal into a perfect differentiator. It will have to compute the derivative of the signal, before noticing that the derivative doesn't exist! So any "differentiator" will be at best an approximation.