In Dummit and Foote, p. 379
Note that any exact sequence can be written as a succession of short exact sequences since to say $X \overset{\alpha}{\to} Y > \overset{\beta}{\to} Z$ is exact at $Y$ is the same as saying that the sequence $0 \to \alpha(X) \to Y \to Y /\ \text{Ker} \beta \to 0$ is a short exact sequence.
What is immediate to me from the second sequence is that there is a bijection between $\text{Im}\ \alpha = \alpha(X)$ and $\text{Im}\ \beta \cong Y /\ \text{Ker} \beta$, but I fail to see the equality $\text{Im}\ \alpha =\ \text{Ker}\ \beta$.
To say: $$0\to\alpha(X)\to Y\to Y/\ker\beta\to0$$Is short exact is precisely to say that $\alpha(X)\to Y$ is injective, $Y\to Y/\ker\beta$ is surjective (I'm assuming some kind of category of rings or modules here) and that the image of $\alpha(X)\to Y$ is equal to the kernel of $Y\to Y/\ker\beta$. Since $\alpha(X)\to Y$ is shorthand for the inclusion map, this is obviously injective, and the quotient $Y\to Y/\ker\beta$ is obviously also surjective (always).
The kernel of $Y\to Y/\ker\beta$ is just $\ker\beta$ and the image of the inclusion $\alpha(X)\to Y$ is just $\alpha(X)$. So this sequence is short exact iff. $\alpha(X)=\ker\beta$, i.e. $\mathrm{im}\,\alpha=\ker\beta$.
So this sequence is short exact iff. the original sequence is exact (at $Y$).