I will focus my answer on the properties which are true for the finite measure spaces but not $\sigma$-finite ones.
Recall Egoroff's theorem:
Let $(X,\mathcal A,\mu)$ a finite measured space, and $\{f_n\}$ a sequence of measurable function from $X$ to $\mathbb R$ endowed with the Borel $\sigma$-algebra. If $f_n\to 0$ almost everywhere then for each $\varepsilon>0$ we can find $A_{\varepsilon}\in\mathcal A$ such that $\mu(X\setminus A_{\varepsilon})\leq\varepsilon$ and $\sup_{x\in A_{\varepsilon}}|f_n(x)|\to 0$.
It's not true anymore if $(X,\mathcal A,\mu)$ is not finite. For example, if $X=\mathbb R$, $\mathcal A=\mathcal B(\mathbb R)$ and $\mu=\lambda$ is the Lebesgue measure, taking $f_n(x)=\begin{cases}1&\mbox{ if }n\leq x\leq n+1,\\\
0&\mbox{otherwise},
\end{cases}$ we can see that $f_n\to 0$ almost everywhere, but if $A$ is such that $\lambda(\mathbb R\setminus A)\leq 1$, then $\mu(A)=+\infty$, hence $A\cap [n,n+1]$ has a positive measure for infinitely many $n$, say $n=n_k$, so $\sup_A|f_{n_k}|\geq \sup_{A\cap [n_k,n_k+1]}|f_{n_k}|=1$.
An explanation could be the following: if $(X,\mathcal A,\mu)$ is $\sigma$-finite, $\{A_n\}$ is a partition of $X$ into finite measure sets, and a sequence converges almost everywhere on $X$, then we have the convergence in measure on each $A_n$: for $k$ and for a fixed $\varepsilon>0$ we can find a $N(\varepsilon,k)\in\mathbb N$ such that $\mu(\{|f_n|\geq \varepsilon\}\cap A_k)\leq \varepsilon$ if $n\geq N(\varepsilon,k)$. The problem, as the counter-example show, is that this $N$ cannot be chosen independently of $k$.
An other result:
Let $(X,\mathcal A,\mu)$ a finite measured space, and $\{f_n\}$ a sequence which converges almost everywhere to $0$. Then $f_n\to 0$ in measure.
We can use the same counter-example as above.
Inclusions between $L^p$ space may change whether the measured space is finite. If $(X,\mathcal A,\mu)$ is a finite measured space, then for $1\leq p\leq q\leq \infty$ we have $L^q(X,\mathcal A,\mu)\subset L^p(X,\mathcal A,\mu)$, as Hölder's inequality shows. But with $X=\mathbb N$, $\mathcal A=2^{\mathbb N}$ and $\mu$ the counting measure, we have for $1\leq p\leq q\leq \infty$, $\ell^p\subset l^q$, so the inclusions are reversed.
I asked this question four months ago. Now that I return to it, many doubts that I had then have disappeared. So I'll write a brief answer below, an answer that I wished to get.
$\sigma$-finiteness is important, as many theorems depend on this property. Non-$\sigma$-finite measure may be too pathological in some sense, but they cannot be avoided altogether. The situation is a bit like Hausdorffness in general topology; most topologies that arise in practice are Hausdorff, but still some authors don't impose this condition.
S. Lang's Real and Functional Analysis treats Bochner integrals right from the beginning (without assuming $\sigma$-finiteness or completeness), and his treatment is almost identical to that of Amann. However, once the usual theory of Lebesgue integration of real-valued functions is established, we can derive properties of Bochner integrals therefrom. Actually, vector-valued integration is typically treated in functional analysis, where Bochner integral is the strong kind and there is also a weak kind called Gelfand–Pettis integral. Maybe that should be the right context for this topic.
I hope this is helpful for people who have the same doubts when reading the text by Amann & Escher.
Best Answer
Consider ${\Bbb R}^2$ with a $\sigma$-algebra $\Sigma$ generated by smooth curves of finite length. Construct a measure $\mu$ so that $\mu(\gamma) = $length of $\gamma$ if $\gamma$ is a curve. This is a measure which is not $\sigma$-finite. The importance of this example is that the map $\mu:\gamma\mapsto\int_{\gamma}d\ell$ is a natural consideration as a measure for an interesting kind of subset (curves) in ${\Bbb R}^2$. In general, on any $n$-dimension manifold, the measure $A\mapsto \int_{A}\omega$ is not $\sigma$-finite if $\omega$ is a differential form or a pseudo differential form of degree $k<n$.