How to prove that End(R) is isomorphic to R? I tried to prove this: 
$R$ as a ring is isomorphic to $End(R)$, where $R$ is considered as a left $R$-module. The map of isomorphism is
$$F:R\to End(R), \quad   F(r)=fr,$$ 
where $fr(a)=ar$. And I define the multiplication in $End(R)$ by $(.)$, where $h.g=g\circ h$ for $g,h$ in $End(R)$. 
Is this true?

With your definition of multiplication in $\def\End{\operatorname{End}}\End(R)$ the map is indeed an isomorphism.
Let's go slowly. We define, for $r\in R$,
$$
F(r)\colon R\to R,\quad x\mapsto xr
$$
It's probably better to use $F_r$ instead of $F(r)$, so
$$
F_r(x)=xr
$$
It's clear that $F_r$ is an $R$-module homomorphism; also
$$
F_r+F_s=F_{r+s}
$$
is a trivial verification.
Let's see what happens with $F_{rs}$:
$$
F_{rs}(x)=x(rs)=(xr)s=F_s(xr)=F_s\circ F_r(x)
$$
for all $x\in R$, therefore
$$
F_{rs}=F_s\circ F_r = F_r\cdot F_s
$$
We have proved that $F\colon R\to\End(R)$ is a ring homomorphism, as $F_1$ is clearly the identity.
Is it an isomorphism? If $F_r$ is the zero map, then $F_r(1)=0$, so $0=r1=r$. Thus $F$ is injective.
If $f\colon R\to R$ is a module homomorphism, set $r=f(1)$. Then
$$
F_r(x)=xr=xf(1)=f(x1)=f(x)
$$
We can push $x$ inside $f$ because $f$ is a homomorphism.

A common convention in module theory is to write morphisms of left modules on the right, but composing them "as written": so, if $f\colon L\to M$ and $g\colon M\to N$ are left module homomorphism, their composition is
$$
f\bullet g\colon L\to N,\quad (x)f\bullet g=((x)f)g
$$
With this convention, your map $F$ becomes an isomorphism between $R$ and $\End(R)$ without changing the operation on $\End(R)$: it's just the same, actually, but computations are more straightforward.
