Show that the sequence is exact We have that $R$ is a commutative ring. 
Suppose that $0\rightarrow A\rightarrow B\overset{f}{\rightarrow} C\rightarrow 0$ and $0\rightarrow C\overset{g}\rightarrow D\rightarrow E\rightarrow 0$ are exact sequences. 

I want to show that $0\rightarrow A\rightarrow B\overset{gf}{\rightarrow} D\rightarrow E\rightarrow 0$ is also exact. 

Could you give me some hints? 

Let $\dots\rightarrow A_{i-2}\overset{f_{i-2}}{\rightarrow}A_{i-1}\overset{f_{i-1}}{\rightarrow}A_i\overset{f_i}{\rightarrow}A_{i+1}\rightarrow\cdots$ be a sequence of $R$-modules and $R$-homomorphisms. 
It is exact at $A_i$ iff $\ker f_i=\text{Im}f_{i-1}$. 
It is exact iff it is exact at each $A_i$. 


What does $0\rightarrow A\rightarrow B\overset{f}{\rightarrow} C\rightarrow 0$ mean? Maybe $0\overset{f}{\rightarrow} A\overset{f}{\rightarrow} B\overset{f}{\rightarrow} C\overset{f}{\rightarrow}0$ ? 

$$$$ 
EDIT: 
We have that each exact sequence can be arised by short exact sequences as above, right? 
But how could we prove this? Could you give me a hint?
 A: I think you want the first term in your exact sequence to be a $0$ (zero )too, not an $O$.
$$0\rightarrow A\rightarrow B\overset{f}{\rightarrow} C\rightarrow 0$$
Just means that we don't bother to give a name to the morphism $A\to B$. In particular it does not mean
$$0\overset{f}{\rightarrow} A\overset{f}{\rightarrow} B\overset{f}{\rightarrow} C\overset{f}{\rightarrow}0$$
Which doesn't even make sense unless $A=B=C=0$.
For the main question, just note that we a map
$$f:B\to C$$
$$g:C\to D$$
Which we can compose to get
$$g\circ f:B\to D$$
You are asked to show that this map makes the longer sequence
$$0\rightarrow A\rightarrow B\overset{g\circ f}{\rightarrow} D\rightarrow E\rightarrow 0$$
exact

I'm not exactly sure what you mean with your new question. Every long exact sequence can be 'weaved together' from short exact sequences as follows:

I'll leave it to you to figure out how to define the $C_i$ in terms of the $A_i$ and maps between them. Let me know if this answers your question or if you are looking for something else.

I'll give some more information:
If we are given any exact sequence $S$:
$$\dots\to A_{n-1}\stackrel{f_{n-1}}{\to}A_{n}\stackrel{f_{n}}{\to}A_{n+1}\stackrel{f_{n+1}}{\to}\dots $$
Then we can construct a sequence of short exact sequences $S_n$ as follows:
The middle term of $S_n$ will be $A_n$. Given the exact sequence $S$, what is a natural object + map that injects into $A_n$? The answer is $\ker f_n$ with the inclusion. What is a natural object and map that $A_n$ surjects into? The answer is $\text{image}(f_n)$ with the map $f_n$. So we construct a new short exact sequence:
$$0\to \ker f_n\stackrel{i}{\to} A_n\stackrel{f_n}{\to}\text{image}f_n\to 0$$
The question is, is this sequence indeed exact? The only non trivial part is checking whether $\text{image}i=\ker f_n$. But this is true by definition, since $i$ is just the inclusion of $\ker f_n$ into $A_n$.
Now we note that if we consider $S_n$ and $S_{n+1}$ we have
$$0\to \ker f_n\stackrel{i}{\to} A_n\stackrel{f_n}{\to}\text{image}f_n\to 0$$
$$0\to \ker f_{n+1}\stackrel{i}{\to} A_{n+1}\stackrel{f_{n+1}}{\to}\text{image}f_{n+1}\to 0$$
But since the sequence $S$ is exact by assumption, $\text{image}f_n=\ker f_{n+1}$. Hence we are in the situation of your original question, and we find that we can combine $S_n$ and $S_{n+1}$ into a longer exact sequence by composing (as in your original question):
$$0\to \ker f_n\stackrel{i}{\to} A_n\stackrel{i\circ f_n}{\to}A_{n+1}\stackrel{f_{n+1}}{\to}\text{image}f_{n+1}\to 0$$
Note that $i\circ f_n=f_n$, hence we can rewrite the above simply as
$$0\to \ker f_n\stackrel{i}{\to} A_n\stackrel{f_n}{\to}A_{n+1}\stackrel{f_{n+1}}{\to}\text{image}f_{n+1}\to 0$$
And we can keep repeating this procedure; this longer sequence can be combined with $S_{n+2}$ and $S_{n-1}$.
In this way every exact sequence arises as a composition of short exact sequences.
A: I name the injective and surjective homomorphisms as $$0\rightarrow  A\overset{j}\rightarrow B\overset{f}{\rightarrow} C\rightarrow 0$$ and $$0\rightarrow C\overset{g}\rightarrow D\overset{p}\rightarrow E\rightarrow 0.$$  

Note that by exactness of 1st sequence, $\operatorname {ker} j=\operatorname {im} (0\to A)=0,$ so that $j$ is injective. By exactness of 2nd sequence, $\operatorname {im} p=\operatorname {ker} (E\to 0)=E,$ so that $p$ is surjective.

To show that 
$0\rightarrow A\overset{j}\rightarrow B\overset{gf}{\rightarrow} D\overset{p}\rightarrow E\rightarrow 0$  is exact one need to see $j$ is injective and $p$ is surjective (which inherites from first and second exact sequences), and
1. $\operatorname {ker}  gf =\operatorname{im} j$
2. $\operatorname{im} gf =\operatorname{ker} p$    


*

*$\operatorname {ker}  gf \overset{def.}= \{x\in B| gf(x)=0\}\overset{g:inj.}=\{x\in B| f(x)=0\}=\operatorname{ker} f = \operatorname{im} j$  

*$\operatorname{im} gf \overset{f:surj.}= \operatorname{im} g =\operatorname{ker} p$


The long exact sequence $\dots\rightarrow A_{i-2}\overset{f_{i-2}}{\rightarrow}A_{i-1}\overset{f_{i-1}}{\rightarrow}A_i\overset{f_i}{\rightarrow}A_{i+1}\rightarrow\cdots$ made by the exact sequences like:
$\dots\rightarrow A_{i-2}\overset{f_{i-2}}{\rightarrow}A_{i-1}\overset{p}{\rightarrow}im f_{i-1}\to 0$ 
and
$0\to im f_{i-1}\overset{j}{\rightarrow}A_i\overset{f_i}{\rightarrow}A_{i+1}\rightarrow\cdots$
