How is it distinguished in matrix multiplication which is the vector and which is the matrix representing a linear transformation? The terminology that is used everywhere when applying a matrix to a "vector" is considered is this: the matrix represents a linear transformation and there is a row or column vector. But a matrix can also be a vector from a vector space because that's the definition for vectors. What happens when matrix multiplication is considered? Can someone explain this mess?
 A: Matrices transform coordinate vectors.  That is, if you want to consider linear transformations on the vector space of $n \times m$ matrices, you should


*

*choose a basis for your vector space

*represent the elements of your vector space (in this case $n\times m$ matrices) as coordinate vectors AKA column matrices

*represent your linear transformation as a $k\times (nm)$ matrix

*perform your matrix multiplication as usual



Let's look at an example:
Say your space is $2\times 2$ matrices with real entries.  This is a $4$-dimensional vector space.  Let's choose an easy basis for this, how about
$$e_1 = \pmatrix{1 & 0 \\ 0 & 0} \\ e_2 = \pmatrix{0 & 1 \\ 0 & 0} \\ e_3 = \pmatrix{0 & 0 \\ 1 & 0} \\ e_4 = \pmatrix{0 & 0 \\ 0 & 1}$$
Now say you are considering the linear transformation $T$ defined by $T(ae_1 + be_2 + ce_3 + de_4) = \pmatrix{a+b & b-c-d \\ c & 0}$.
Then the matrix $M$ which represents this transformation should always obey the equation:
$$M\pmatrix{a \\ b \\ c \\ d} = \pmatrix{a+b \\ b-c-d \\ c \\ 0}$$
Then we can see by the linearity of matrix multiplication that this is just 
$$aM_1 + bM_2 + cM_3 + dM_4 = a\pmatrix{1 \\ 0 \\ 0 \\ 0} + b\pmatrix{1 \\ 1 \\ 0 \\ 0} + c\pmatrix{0 \\ -1 \\ 1 \\ 0} + d\pmatrix{0 \\ -1 \\ 0 \\ 0}$$
where $M_i$ is the $i$th column of the matrix $M$.  Thus
$$M = \pmatrix{1 & 1 & 0 & 0 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0}$$
And finally if you want to use matrix multiplication to find out how $M$ transforms the matrix $x = \pmatrix{1 & 1 \\ 1 & 1}$, then just multiply $M$ by the coordinate vector of the matrix $x$:
$$M[x] = \pmatrix{1 & 1 & 0 & 0 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0}\pmatrix{1 \\ 1 \\ 1 \\ 1} = \pmatrix{2 \\ -1 \\ 1 \\ 0}$$
Therefore $$\color{red}{T(x) = \pmatrix{2 & -1 \\ 1 & 0}}$$

I should also note, though, that matrix multiplication can have a different interpretation as well.  Multiplying two matrices results in a matrix that represents the composition of linear transformations.
For instance, say $S: \Bbb R^3 \to \Bbb R^2$ and $T: \Bbb R^3 \to \Bbb R^3$ are two linear transformations.  Further let $A$ be the matrix which represents $S$ in the basis we're concerned with and likewise $B$ be the matrix representing $T$.
Then what's the matrix which represents the composition of these two linear transformations, $S \circ T$?  Well it's just $AB$.
This is easy to show.  By definition $S(x) = Ax$ and $T(y) = By$, therefore $$(\color{red}{S \circ T})(y) = S(T(y)) = S(By) = A(By) = (\color{red}{AB})y$$
A: If I understand correctly, you want to know where multiplication of matrices, by matrices, stands when considering a certain space $\mathcal M$ of matrices as a vector space. In this case you simply have a vector space over $\Bbb R$, whose elements happen to form a ring. In fact you have an $\Bbb R$-algebra.
Another example of this is the space of continous functions $\mathcal C(\Bbb R,\Bbb R)$. It is an $\Bbb R$ vector space, but function multiplication is also defined by $(fg)(x)= f(x)g(x)$. This operation also makes it an $\Bbb R$-algebra.
