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.
Hint
Denote $(X^n)^+$ and $(X^n)^-$ the positive and negative part. One can prove that if $n$ is odd, then $(X^n)^+=(X^+)^n$ and $(X^n)^-=(X^-)^n$. And if $n$ is even, $(X^n)^+=X^n$ and $(X^n)^-=0$. Then,
\begin{align*}
\mathbb E[(X^+)^n]&=\int_{\Omega } (X^+(\omega ))^n\mathbb P(\mathrm d \omega )\\
&=\int_\Omega \int_0^{X^+(\omega )}nx^{n-1}\,\mathrm d x \mathbb P(\mathrm d \omega )\\
&\underset{\text{Fubini}}{=}\int_{0}^\infty nx^{n-1}\int_\Omega \boldsymbol 1_{\{X(\omega )>x\}}\mathbb P(\mathrm d \omega )\,\mathrm d x\\
&=n\int_0^\infty x^{n-1}\mathbb P\{X(\omega )^+>x\}\,\mathrm d x.
\end{align*}
I let you do the other cases and conclude.
Best Answer
From Fubini's theorem existos a $y_0$ such that $f(x)-f(y_0)$ is $\mu-$ integrable
That is $\int |f(x)-f(y_0)| d\mu(x)< \infty$
So $$\int|f(x)|d\mu(x) \leq \int|f(x)-f(y_0)| d\mu(x)+\int|f(y_0)| d\mu(x)$$ $$\int|f(x)-f(y_0)| d\mu(x)+|f(y_0)|<\infty$$