What is the underlying difference between applying linear operator as $xA$ and $Ax$? In different sources on linear algebra I meet different forms of linear transformation: $Ax$ (1) and $xA$ (2), where 
$\newcommand{\RR}{\mathbb{R}}A:\RR^{n}\rightarrow \RR^{n}$ - linear operator
$x \in \RR^{n}$ - vector in $\RR^n$
I understand, that in situtations (1) and  (2) matrix multiplication is being done in different order.
The question is: how to interpret this difference geometricly, physically?
 A: $Ax $ is a column vector that corresponds to a columns of A combination by $x_i$ coefficients
$x^TA=(A^Tx)^T$ is a row vector that corresponds to a rows of A combination by $x_i$ coefficients
A: The terms "column vector" and "row vector" apply to matrix algebra, not linear algebra.  In linear algebra both "Ax" and "xA" are different notations for "A(x)", the linear transformation (which is a function) applied to the vector x.
A: There's a couple points to address here. First I'll say that in some ways $xA$ and $Ax$ are fundamentally different, and $x$ must be different as well between the two cases. I'll first explain this with matrices before moving into linear algebra.
As you may know, you can multiply a $n\times m$ matrix $A$ and a $n'\times m'$ matrix $B$ to get a $n\times m'$ matrix $AB$ exactly when $m=n'$. So if $x$ is "vector-like", i.e., one of its dimensions is 1, and $Ax$ is defined, $x$ must have the shape $m\times 1$, so $x$ is a column-vector of height $m$. Conversely if $xA$ is defined, $x$ must have the shape $1\times n$, so $x$ is a row-vector of length $n$. Also if $Ax$ is defined, it has shape $n\times 1$, so $Ax$ is a column-vector of height $n$. Conversely, in the other order, if $xA$ is defined, it has shape $1\times m$, so it is a row-vector of length $m$. The key things to notice are that unless $A$ is square, multiplying on the left and right require different length vectors, so they should be fundamentally different somehow. This should be true even in your case when they are square. The other key point is that $A$ expects $x$ to be a column-vector for $Ax$ and a row-vector for $xA$, and this is true even when $A$ is square, as it is in your case.
Now I don't think matrix algebra is particularly illuminating when we want to describe what's going on here and get intuition. It only helps us formalize things to tell us where to look. So let's move into linear algebra, and see if it can help us explain what we noticed about the matrices. 
Let $\newcommand{\RR}{\mathbb{R}}A:\RR^m\to \RR^n$ (this is the equivalent to a $n\times m$ matrix). Then if $x\in\RR^m$, $Ax\in\RR^n$. This mirrors the observation in the matrix algebra that if $x$ is a column-vector of height $m$, $Ax$ should be a column-vector of height $n$. Now what is a row-vector? If you think about the correspondence between matrices and linear transformations, you'll see that a row-vector is a linear map $\alpha: \RR^n\to \RR$ (I'm going to switch to $\alpha$ instead of $x$ for hopefully greater clarity).
Then the composition, $\alpha \circ A$, usually written $\alpha A$ is a linear map from $\RR^m\to \RR$. This mirrors the observation above that if $x$ was a row-vector of length $n$, then $xA$ should be a row-vector of length $m$. 
Side note on notation, if $V$ is a vector space, the collection of linear maps from $V$ to $\RR$, is called the dual space of $V$, and is often denoted by $V^*$. It's not so hard to check that this is itself a vector space. Thus we could say above that $\alpha\in\RR^{n*}$ for brevity. 
Now if $m=n$, corresponding to the case where $A$ is a square matrix, there ought to be some way to take a vector in $\RR^n$ and get a linear map $\RR^n\to \RR$. In matrix algebra, this is the transpose, since transpose takes column-vectors to row-vectors, so we want to understand what this is in linear algebra. Well, it turns out that we need to remember we have the dot product. So we take a vector $x$ and send it to the linear map $y\mapsto x\cdot y$. I will call this map $x^T$, to keep it in line with the matrix notation. Thus $x^T A$ is the linear map that takes in a vector $y$, sends it to $Ay$, then dots it with $x$, i.e. the map $y\mapsto x\cdot Ay$.
Ok, now we have the linear algebraic picture, which should directly translate into geometry. I assume you're familiar with how to visualize vectors, but I should probably explain how to visualize dual-vectors. You can think of $\alpha\ne 0\in\RR^{n*}$ as describing a hyperplane. The kernel of $\alpha$ will be a hyperplane in $\RR^n$, and you can think of $\alpha$ as being the normal to this hyperplane. $\alpha x$ describes how close $x$ is to the hyperplane $\alpha$. If $\alpha$ is zero, it doesn't describe a hyperplane, it just sends everything to 0. Now if $x$ is a vector, the corresponding dual vector $x^T$ describes the hyperplane perpendicular to $x$.
Alright now we can put it all together. $Ax$ is easy, it's just the vector $A$ sends $x$ to. On the other hand, $x^TA$ should be a hyperplane, but which hyperplane? It's the hyperplane of all vectors that $A$ sends into the hyperplane perpendicular to $x$. It's the preimage of the hyperplane perpendicular to $x$ under $A$. You do have to be a bit careful here, $x^TA$ won't be a hyperplane if $x^TA=0$, which happens when $A$ sends every vector into the hyperplane perpendicular to $x$. 
Alright, hopefully that was helpful.
A: It's simply a difference of notation.  Geometrically you have vectors acted upon by linear operators.  In a finite-dimensional case you may choose to write the vectors as columns, and the operators as matrices acting on the left of them ($Ax$), or you may choose to write the vectors as rows, and the operators as matrices acting on the right ($xA$).  These are related by $(Ax)^T = x^T A^T$.  
