for simplicity a specific case:
$f:=(a_i)_{i \in \mathbb{N}} \in \mathbb{R}[X]$
$A \in \mathbb{R}^{4\times 4}$
In one of our assignments f is evaluated as $\sum_{i=0}^{\deg(f)} a_iA^i$.
How is this evaluation possible formally since $\mathbb{R}$ is not a subring of $\mathbb{R}^{4 \times 4}$? Reference: https://en.wikipedia.org/wiki/Polynomial_ring#Polynomial_evaluation
greets
Ultimately, what matters is that the operations make sense after substitution.
If you take some structure that is not a ring, say the natural numbers $\mathbb{N}$, and you try to compute $f(a)$ for $a$ in $\mathbb{N}$, you probably won't find an integer.
This is why, in the article, the element $a$ has to be part of a ring.
Now, why does $R$ have to contain $K$? This is to allow the multiplication by the $a_i$ coefficients of the polynomial.
Take the ring $M_4(\mathbb{Z})$, matrices with integers coefficients, and $M$ a matrix in that set. Then, what is $\pi M$? It is not part of our ring $M_4(\mathbb{Z})$.
In our case, you can multiply a matrix that has real coefficients by a real number, and get back a matrix with real coefficients (some call this an algebra).
You can also add two real matrices together, and get back a real matrix. Thus, all the operations make sense.
Another way to answer your question would be to say that $$\tilde K = \{\lambda I_4, \lambda \in \mathbb{R}\}$$ is isomorphic to $(\mathbb{R}, +, .) = K$. Indeed, the function $$\lambda \in \mathbb{R} \mapsto \lambda I_4 \in \tilde K$$ is a ring isomorphism, mapping $K$ to $\tilde K$. So, in a way, $M_4(\mathbb{R})$ contains $\mathbb{R}$.
Another interesting property to think about is commutativity. Indeed, matrices do not commute in general. However, if you stick to one matrix, and transformations such as $(+, .)$ applied iteratively to the $a_iA^k$, this works just fine.