Your Professor was probably talking about Tonelli's Theorem in regard to $\sigma$-finiteness.
If $f \in L^{1}(\mu\times\nu)$, then Fubini's theorem holds, regardless of $\sigma$-finiteness of $\mu$, $\nu$ or not. Of course all of the measures must be complete, including the product measure. The way this is proved is by reducing to the case of positive $f$ because the positive $f_{+}$ and negative parts have finite integrals. That allows you to approximate $f_{+}$, for example, by a non-decreasing sequence of non-negative simple functions $\{\varphi_{n}\}$ converging upward to $f_{+}$ with the property that each is supported on a set of finite measure. This approximation is a critical part of the standard proof.
Tonelli's Theorem is a generalization of Fubini's Theorem for the case of positive functions $f$, where the assumption of integrability of $f$ is dropped; that is, you allow for the possibility that $\int f\,d(\mu\times\nu) = \infty$. You still get the same conclusion as Fubini's Theorem for such a case, provided you assume that the measures $\mu$ and $\nu$ are $\sigma$-finite. By adding this assumption of $\sigma$-finiteness, you are still able to get the existence of $\{\varphi_{n}\}$ as above which are once again supported on sets of finite measure. So the proof of Fubini's Theorem goes through, even without assuming $f$ has a finite integral. However, in this case, I think you can see the need for $\sigma$-finiteness, whereas in Fubini's Theorem, it was only necessary to assume $\int |f|\,d(\mu\times \nu) < \infty$ in order to get the desired approximation.
Suppose $\mu_1=\mu_2$ are counting measures on $\Omega_1=\Omega_2=\{1,2,\ldots\}$.
Define the following function on $\Omega_1\times\Omega_2$:
$$f(i,j)=\begin{cases}1&,\text{ if }i=j
\\ -1&,\text{ if }i=j+1
\\ 0&,\text{ otherwise }
\end{cases}$$
We can write out the values of $f(i,j)$ in a matrix form like
$$[f(i,j)]=\begin{bmatrix}1&0&0&0&\cdots
\\ -1&1&0&0&\cdots
\\0&-1&1&0&\cdots
\\0&0&-1&1&\cdots
\\\vdots&\vdots&0&-1&\cdots
\\\vdots&\vdots&\vdots&\vdots&\ddots
\\0&0&0&0&\cdots
\end{bmatrix}$$
Only the first row sums to $1$, each of the remaining rows sum to $0$. Also sum of each column is $0$.
Therefore, $$\int\left(\int f(x,y)\,d\mu_2(y)\right)d\mu_1(x)=\sum_{i=1}^\infty \left(\sum_{j=1}^\infty f(i,j)\right)=1$$
And $$\int\left(\int f(x,y)\,d\mu_1(x)\right)d\mu_2(y)=\sum_{j=1}^\infty \left(\sum_{i=1}^\infty f(i,j)\right)=0$$
However,
\begin{align}
\iint|f(x,y)|\,d\mu_1(x)\,d\mu_2(y)&=\sum_{i=1}^\infty\sum_{j=1}^\infty|f(i,j)|
\\&=\sum_{i=1}^\infty\left(\sum_{j=1}^\infty |f(i,j)|\right)\quad,\small\text{ by Fubini/Tonelli, since }|f|\ge 0
\\&=1+2+2+\cdots
\\&=\infty
\end{align}
So $f$ is not $\mu$-integrable where $\mu=\mu_1\otimes\mu_2$ is the product measure.
Best Answer
Wikipedia has actually an alternate theorem statement that answers the question. Besides $\sigma$-finiteness, both iterated integrals of the absolute value of the function have to be finite.
Now $\sigma$-finiteness is implicitely required in Fubini's theorem to some degree. The assumption $$\int_{A\times B}|f(x,y)|d(x,y) < \infty.$$ implies that $F_n=\{(x,y):|f(x,y)|>1/n\}$ has finite measure, so the product measure restricted to $\bigcup_n F_n=\{(x,y):f(x,y)\neq 0\}$ is $\sigma$-finite.