Short exact sequences of modules. My understood definition: $0 \to A\stackrel{\psi} \to B\stackrel{\phi} \to C \to 0$ is a short exact sequence of modules iff $\psi$ is injective, $\phi$ is surjective, and $\ker \phi = \psi(A)$, which is equivalent to $B/A \approx C$.  Is that correct?
And also how do you arrive at $A\stackrel{\psi} \to B\stackrel{\phi}\to C$ is exact iff $0 \to \psi(A) \to B \to B/\ker \phi \to 0$ is short exact?
 A: For the first part, the general definition is that a sequence $\cdots \to L \xrightarrow{f} M \xrightarrow{g} N \to \cdots $ is exact at $M$ if and only if the image of $f$ is the kernel of $g$.
A short exact sequence is an sequence consisting of five terms whose endpoints are zero, and is exact at every (internal) point. i.e.
$$ 0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0 $$
is exact at $A$, at $B$ and at $C$. Exactness at $A$ means that the kernel of $f$ is the image of the zero map: i.e. it is $\{ 0 \}$, so exactness at $A$ is equivalent to $f$ being injective. Similarly for exactness at $C$.
Usually it doesn't even make sense to say $B/A$, because $A$ usually isn't a submodule of $B$! However, there is equivalence in the sense that there is an isomorphism of short exact sequences:
$$ \begin{matrix}
0 &\to & A &\xrightarrow{f}& B &\xrightarrow{g}& C &\to& 0 
\\  & & \ \downarrow^f & & \ \downarrow^1 & & \downarrow
\\ 0 &\to & f(A) &\xrightarrow{i}& B &\xrightarrow{\pi}& B / f(A) &\to& 0 
\end{matrix}
$$
where both rows are short exact sequences, $i$ and $\pi$ are the inclusion and projection morphisms, all three vertical morphisms are isomorphisms, and the diagram is commutative. (meaning, e.g., if you take all paths from the top $A$ to the bottom $B$ and compose the arrows, you get the same result: $i \circ f = 1 \circ f$)
One can view that the situation as per the definition of quotient module is somewhat deficient, and the idea of a short exact sequence is what one really should be thinking about.
A: You need to be a little more careful in your first claim. Consider the short exact sequence
$$
0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/n \to 0
$$
where the injection $\phi$ is multiplication by $n$ and the surjection is the natural quotient map. What you're written is that $\mathbb{Z}/n \simeq \mathbb{Z} / \mathbb{Z} = 0$ but what is actually true is that $\mathbb{Z}/n \simeq \mathbb{Z} / im(\phi) = \mathbb{Z} / n\mathbb{Z}$. So although $im(\phi) \simeq \mathbb{Z}$, we can't carry this through the quotient. 
For your second question, just use the definitions. We need $\psi$ to be injective, $\phi$ to br surjective and $im \psi = \ker \phi$. Can you see how this gives the if and only if?
A: About your first question: yes, that is the definition of a short exact sequence. 
For the second question, note that if $\phi: B \rightarrow C$ is any $R$-map then
$$
0 \rightarrow \text{ker}(\phi) \rightarrow B \rightarrow \text{im} (\phi) \rightarrow 0 
$$
is short exact. In particular,  $\text{im}(\phi) \cong B / \text{ker}(\phi)$. Now, suppose that
$$  
A \rightarrow^{\psi} B \rightarrow^{\phi} C 
$$
is exact then ker$\phi =\text{im}\psi$. So, using our first sequence we get that
$$
0 \rightarrow \text{im}(\psi) \rightarrow B \rightarrow B/ \text{ker}(\phi) \rightarrow 0
$$
is short exact. Conversely, if 
$$
0 \rightarrow \text{im}(\psi) \rightarrow B \rightarrow B/ \text{ker}(\phi) \rightarrow 0
$$
is short exact then $\text{im}(\psi) \rightarrow B$ is injective and so is an isomorphism onto its image. By exactness of the sequence this image is $\text{ker}(\phi)$. 
A: Let $X \xrightarrow{\alpha} Y \xrightarrow{\beta} Z$ be exact.  We know that $\iota: \alpha(X) \hookrightarrow Y$ is injective and $\pi : Y \to Y/\ker{\beta}$ is surjective.  So that $0 \to \alpha(X) \to Y \to Y/\ker \beta \to 0$ is a "short sequence".  But since $\alpha(X) = \ker \beta$, we have that this sequence is clearly exact as well.  So it is a short exact sequence.
Conversely, let $\alpha, \beta$ be two morphisms such that $0 \to \alpha(X) \to Y \to Y/\ker \beta \to 0$ is short-exact, then $\alpha(X) = \ker \beta$ so the first diagram is exact.  We're done.
