Well, $V_i$ is an arbitrary object of the putative exact sequence, so if you checked exactness there, that means you're done.
Or, that means you're done checking that the sequence of $V_i$ is an exact sequence, which is not quite what you set out to prove.
You set out to prove that such an exact sequence of $V_i$ "is the same as having a collection of short ones such that...", and yet you merely showed one way: if you have a collection of short exact sequences, you can splice them up.
Now you must show that given a long exact sequence you can chop it up in short exact sequences.
To see that, a diagram is worth a thousand words:
where you define $K_n=\operatorname{ker}(f_n)$.
The map $K_n\to A_n$ is the inclusion of the kernel of $f_n$. To get the map $A_{n+1} \to K_n$, observe that:
$K_n=\operatorname{ker}(f_n)=\operatorname{im}(f_{n+1})=\frac{A_{n+1}}{\operatorname{ker}(f_{n+1})}=\frac{A_{n+1}}{\operatorname{im}(f_{n+2})}=\operatorname{coker}(f_{n+2})$.
Hence, define the map $A_{n+1}\to K_n$ as the quotient map of $A_{n+1}$ by the image of $f_{n+2}$.
It is clear by definition that $0\to K_{n+1}\to A_{n+1}\to K_n\to 0$ are short exact sequences.
$ $
Now, you worry about how to chop a long exact sequence into short exact sequences. You shouldn't worry, since if
is an exact sequence, then you can enlarge it to an exact sequence infinite on both sides:
$ $
I wrote a couple of pages about splicing and decomposing exact sequences some time ago, which has a couple more results that you might find useful. There is also a section on equivalent definitions of exact functors which exploits these considerations. It is here: it's in Spanish, though.
You are right, it finally involves cokernels (=coimages of the next arrows).
Your first line is correct: taking cokernel of both sides of $\def\im{{\rm im}\,} \im f=\ker g$, we'll get $\def\coker{{\rm coker}\,} \def\coim{{\rm coim}\,} \coker f=\coim g$. For the converse, take kernel of both sides.
Then, for the objects, your second line should read
$${\rm Im\,}g\cong {{\rm Coker}\,f}\,.$$
The statement ${\rm Coker\,}g\cong{\rm Ker\,}g$ does not follow.
And yes, this is the way how the long exact sequence splits into short exact sequences.
Best Answer
From what I understand, you are only used to the concept of short exact sequences, i.e. sequences of the form
$$ 0\to A\to B\to C\to 0\,, $$
missing the general concept of an exact sequence. The latter is a sequence $(A_i,d_i)$ of modules $A_i$ and homomorphisms $d_i\colon A_i\to A_{i+1}$ (conventions vary in which direction the maps go) such that $\ker(d_{i+1})=\operatorname{im}(d_i)$ for all $i$, which ones refers to as exactness at $A_i$ as the $\ker(d_{i+1}),\operatorname{im}(d_i)\le A_i$. As I do not intend to talk about this in more generality, we may as well assume that this sequence is in fact of the form
$$ 0=A_0\xrightarrow{d_0} A_1\xrightarrow{d_1} A_2\to\cdots\to A_{n-1}\xrightarrow{d_{n-1}} A_n\xrightarrow{d_n} A_{n+1}=0\,. $$
In this case, $\ker(d_{i+1})=\operatorname{im}(d_i)$ dictates that, in particular,
$$ \ker(d_1)=\operatorname{im}(d_0)=0\quad\text{and}\quad\operatorname{im}(d_{n-1})=\ker(d_n)=A_n $$
which includes your observations as special case.
Now suppose we are given two short exact sequences
$$ 0\to A\xrightarrow{\iota} B\xrightarrow{f} C\to 0\quad\text{and}\quad0\to C\xrightarrow{g} D\xrightarrow{\pi} E\to 0 $$
and we want to show that the induced sequence
$$ 0\to A\xrightarrow{\iota} B\xrightarrow{gf} D\xrightarrow{\pi} E\to 0 $$
is then exact too. Exactness at $A$ and $E$ is still given as we use the same maps. So the relevant things to show are exactness at $B$ and at $D$.
The fastest way of doing this is to realize that
$$ \ker(gf)=\ker(f)\quad\text{and}\quad\operatorname{coker}(gf)=\operatorname{coker}(g)\,, $$
which follows from $g$ being a monomorphism and $f$ being an epimorphism (which is equivalent to being injective and surjective but requires some work), and using an alternative characterization (proposition $2.4$) of exactness in terms of cokernels. However, this might be a bit over the top. The main idea stands, nonetheless, i.e. using that $g$ and $f$ are injective and surjective (i.e. mono- and epimorphisms), resp., which then elementwise translates to
\begin{align*} ((g\circ f)\circ\iota)(x)=0\,&\iff\,(f\circ\iota)(x)=0 \\ (\pi\circ(g\circ f))(x)=0\,&\iff\,(\pi\circ g)(x)=0 \end{align*}
from where it should be clear how to apply the exactness of the original sequences.
Can you take it from here? If some terminology or ideas are unclear (mono-/epimorphisms, cokernels, etc.) please let me know. I can try to make the answer more elementary or include an explanation.