What does this notation mean? $\frac{\partial f}{\partial x}(x+y)=\frac{\partial }{\partial x}f(x+y)$ 
$$\frac{\partial f}{\partial x}(x+y)=\frac{\partial }{\partial x}f(x+y)$$

I was just wondering what the left-hand side mean. (or how to do the operation based on the notation of the LHS, given a specific function $f$)
In addition, when is such commutation true? Under what conditions?
 A: If $f$ is some function, then we may define $f_y : x\mapsto f(x+y)$.  It is basically a shifted version of the function $f$. 
Your equality says that (with $D$ the derivation operator):
$$ Df(x+y) = Df_{y}(x), $$
or, if you prefer
$$ (Df)_y = D(f_y). $$
In other words: "shifting" or "translating" a function commutes with taking derivatives. If we denote by $T_y$ the operator $f\mapsto f_y$, then we could write this as
$$T_y\circ D = D\circ T_y $$
In the language of systems one says that derivation is a time-invariant system. In my opinion it's also a good example of how the $\frac{\partial}{\partial x}$  can sometimes obscure things.

This is always true (if the derived function exists), because (intuitively) derivation only depends on a neighbourhood of the function, and not on the ordinate. In other words, if I give you a picture of a function but forget to draw the $y$-axis, you could still sketch the derived function. (This would be different if I asked you to, for instance, draw $xf(x)$.)
For a more rigorous derivation of this property: let $f$ be any (differentiable) function, then we have that
$$ \begin{align*} Df_y(x) &= \lim_{h\to 0}\frac{f_y(x+h)-f_y(h)}{h} \\ &= \lim_{h\to 0} \frac{f(x+h+y)-f(x+y)}{h} \end{align*}$$
and
$$ \begin{align*} (Df)_y(x) &= Df(x+y)\\ &= (Df)(z)|_{z=x+y} \\ &= \lim_{h\to 0}\left.\frac{f(z+h)-f(z)}{h} \right|_{z=x+y} \\ &= \lim_{h\to 0} \frac{f(x+y+h)-f(x+y)}{h} \end{align*} $$
