Clarification regarding linear transformation from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^{2\times 2}$ Every linear transformation has a matrix representation, once we've chosen a basis.
Suppose I define a linear transformation $T:\mathbb{R}^3\rightarrow \mathbb{R}^2$ as follows: $T(x,y,z)=(x+y,y+z)$.
Now, suppose I pick the standard basis (for simplicity, let's do that for the rest of the discussion) for $\mathbb{R}^3$ and $\mathbb{R}^2$, then the representing matrix is:  
$$[T]=\bigg(\begin{matrix}1 & 1& 0\\ 0& 1& 1\end{matrix}\bigg)$$
Indeed, for every $v\in \mathbb{R}^3$, we have $T(v)=[T][v]$.
It is also not too hard to deduce the dimensions of that matrix; the representing matrix should be a $\dim(\mathbb{R}^3)\times\dim(\mathbb{R}^2)=3\times 2 $ matrix, and indeed this is what we have here. I have always used this rule for self checking.  
That's all nice and well, and this is basically what I had to deal with so far, but I'm confused about what exactly is going on when the vectors in the vector spaces are matrices themselves.
For example, take the following linear transformation: $T:\mathbb{R}^{2\times 2}\rightarrow \mathbb{R}^{2\times 2}$  defined by:
$$T\bigg(\begin{matrix}w & x\\ y& z\end{matrix}\bigg)=\bigg(\begin{matrix}w+y & x+z\\ y& z\end{matrix}\bigg)$$
In one hand, it is easy to see that $A=\bigg(\begin{matrix}1 & 1\\ 0& 1\end{matrix}\bigg)$ is the representing matrix (that is, $[T]=A$).
Namely, for every $M\in \mathbb{R}^{2\times 2}$, it holds that $T(M)=AM$.
But on the other hand, $\dim(\mathbb{R}^{2\times 2})\times \dim(\mathbb{R}^{2\times 2})\neq 2\times 2$.
Why doesn't this rule apply here? am I missing something here, or that's just the way it should be?

If you write an element of $\Bbb R^{2\times 2}$ as a $2 \times 2$ matrix, then there are many linear transformations that cannot be represented as a $2\times 2$ matrix, for instance taking the transpose:
$$
\begin{pmatrix}x& y\\z & w\end{pmatrix} \mapsto \begin{pmatrix}x& z\\y & w \end{pmatrix}
$$
This lack of representability is closely linked to this lack of dimensions that you've discovered.
What you need to do is interpret $\Bbb R^{2\times 2}$ as $\Bbb R^4$. Then any linear transformation $\Bbb R^{2\times 2} \to \Bbb R^{2\times 2}$ can be represented by a $4 \times 4$ matrix.
