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I am having trouble understanding the following lemma:

Let $f: M \to N$ be an $A$-module homomorphism. Then there is an exact sequence $$0 \rightarrow K' \stackrel{i}{\rightarrow} M \stackrel{f}{\rightarrow} N \rightarrow C' \stackrel{\pi}{\rightarrow} 0,$$ where $i$ is the inclusion of the kernel of $f$ and $\pi$ is the canonical map onto the cokernel. Conversely, given a six term exact sequence $$0 \rightarrow K' \rightarrow M \stackrel{f}{\rightarrow} N \rightarrow C' \rightarrow 0,$$ $f$ is injective iff $K'=0$ and surjective iff $C'=0$.

I'm having trouble with the converse.First of all, is that six term sequence well-defined? How can it be if there is only one homomorphism involved? Are the homomorphisms suppressed? Should it actually read

$$0 \rightarrow K' \stackrel{g}{\rightarrow} M \stackrel{f}{\rightarrow} N \stackrel{h}{\rightarrow} C' \rightarrow 0?$$ where $g : K' \to M$ and $g : N \to C'$ are homomorphisms. Even with this modification, I'm still having trouble seeing why it is true. Certainly if $K'=0$ and $C'=0$, then $f$ will be bijective. But why should be $K'=C'=0$ if $f$ is bijective?

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First, unlabelled arrows do represent homomorphisms, just that they haven't been given a name. You are free to name them yourself.

Recall that any homomorphism $f$ is injective iff $\ker f = 0$, and surjective iff $\operatorname{coker} f = 0$.

So given any exact sequence $$0 \rightarrow K' \stackrel{g}{\rightarrow} M \stackrel{f}{\rightarrow} N \stackrel{h}{\rightarrow} C' \rightarrow 0,$$ you can show that $K' \cong \ker f$ and $C' \cong \operatorname{coker} f$. For instance, $K' \cong \ker f$ because by exactness, $g$ is injective and $\operatorname{im} g = \ker f$.

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