I am answering my own question because I discovered the answer in my notes, and it may help someone else (it will definitely help me as I return to reference this page in the future).
Here is what is meant by $\sigma$-finiteness being "hidden" in the hypotheses of Fubini's theorem.
To prove Fubini's theorem, we assumed $f \in L^{1}(d\lambda)$. Notice that $f = f\chi_{ \{ x \mid f(x) \neq 0 \} }$, where $\chi_{A} = \begin{cases} 1 & x \in A \\ 0 & x \not \in A \end{cases}$.
Then that means $\int \limits_{X} f \,d\lambda = \int \limits_{X} f\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$. And $\int \limits_{X} f\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{X} f \,d(\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda)$ (for a proof, see the answer and comments here). The measure given by $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$ is $\sigma$-finite, because:
$X = \bigcup \limits_{n = 1}^{\infty} \{ x \mid |f| \geq \frac{1}{n} \} \bigcup \{ x \mid f = 0 \}$.
Now we just need to show each of these sets in the countable union has finite measure with respect to the measure $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$.
For each $n$, we have $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda( \{ x \mid |f| \geq \frac{1}{n} \}) = \int \limits_{ \{ x \mid |f| \geq \frac{1}{n} \} } 1 \chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{ X \times Y } \chi_{\{ x \mid |f| \geq \frac{1}{n} \}} \chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{ X \times Y } \chi_{\{ x \mid |f| \geq \frac{1}{n} \} \cap \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{ X \times Y } \chi_{\{ x \mid |f| \geq \frac{1}{n} \} } \,d\lambda \leq \int \limits_{ X \times Y } n |f| \,d\lambda < \infty $
since $f \in L^{1}(d\lambda)$. So for each $n$, $ \{ x \mid |f| \geq \frac{1}{n} \} $ has finite measure with respect to the measure $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$. Also, it's clear that $\{ x \mid f(x) = 0 \}$ has measure $0$ with respect to this measure.
So, $(X, \Sigma, \chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda)$ is $\sigma$-finite.
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
I think that your idea does not work immediately. Indeed, if $f \in L^1(X; F)$, you can take a bounded linear operator $A \colon F \to G$ and have $$ \int_X A f(x) \, \mathrm{d}x = A \int_X f(x) \, \mathrm{d}x. $$ This does not yield your desired formula in case $f \in L^1(X \times Y; E)$, since the integral over $Y$ is not a linear operator on the space $X$.
However, the space $L^1(X \times Y; E)$ is isometrically isomorphic to $L^1(X; L^1(Y;E))$. Let $I \colon L^1(Y;E) \to E$ be the Bochner integral. Then, $$ \int_X \int_Y f \, \mathrm{d}y \, \mathrm dx = \int_X I f \,\mathrm dx = I \int_X f \,\mathrm dx = \int_Y\int_X f \,\mathrm{d}x\,\mathrm{d}y $$ for all $f \in L^1(X; L^1(Y;E))$.
One should be aware that the proof of the mentioned identification of spaces might already require some Fubini-type argument (I do not remember exactly...)