Here is a rather "simple" counter-example using Borel (or Lebesgue) $\sigma$-algebra and the Lebesgue measure $m$. Consider $[0,1]$ and let $\omega_1$ be the first uncountable ordinal. Assuming the Continuum Hypothesis, there is a bijection $o:[0,1] \to \omega_1$.
Define $E=\{(x,y)\in [0,1]^2 : o(x)<o(y)\}$. Then, for all $x\in [0,1]$,
$$E_x = \{y\in [0,1] : o(x)<o(y)\} =[0,1] - \{y\in [0,1] : o(y)\leqslant o(x)\}$$
so $E_x$ is the complement (in $[0,1]$) of a countable subset. So $E_x$ is measurable and $m(E_x)=1$.
Now, for all $y\in [0,1]$,
$$E_y = \{x\in [0,1] : o(x)<o(y)\}$$
so $E_y$ is a countable subset of $[0,1]$. So $E_y$ is measurable and $m(E_y)=0$.
Claim: $E$ is not measurable
Let us prove it by contradiction. Suppose that $E$ is measurable. Then $\chi_E$ is a non-negative measurable function and, by Tonelli's theorem, we have
$$ 0= \int_0^1 \left(\int_0^1 \chi_E(x,y) dx \right ) dy = \int_0^1 \left(\int_0^1 \chi_E(x,y) dy \right ) dx =1$$
Contradiction.
Remark: The set $E$ above is a Sierpinski set.
Remark 2: I assume you was looking for a counter-example using Borel (or Lebesgue) $\sigma$-algebra. Otherwise, there are simpler counter-examples.
Remark 3: You wrote: "By Fubini-Tonelli's theorem, we know that if $E\subset \mathbb{R^{n+m}}$ is measurable
then the sections $E_x=\{y\in \mathbb{R^m}: (x,y)\in E\}$ and $E_y$ are measurable".
This is not true if you are considering the Lebesgue $\sigma$-algebra. Moreover, in this case, Fubini-Tonelli's theorem implies only that $E_x$ and $E_y$ are measurable for almost every $x$ and almost every $y$ (respectively).
Let's write down a statement of Tonelli's theorem, just to make everything clear.
Let $(X, \mathcal M, \mu)$ and $(Y, \mathcal N, \nu)$ be $\sigma$-finite measure spaces and $f: X \times Y \to [0,\infty]$ be $\mathcal M \otimes \mathcal N$ measurable. Then:
$$\int f d(\mu \times \nu) = \int \left(\int f(x_1, x_2) d\nu(x_2)\right) d\mu(x_1) = \int \left(\int f(x_1, x_2) d\mu(x_1)\right) d\nu(x_2).$$
Alright now we can write down a proof of your statement. Let $(X_j, \mathcal M_j, \mu_j)$ be a finite collection of $\sigma$-finite measure spaces and let $f : \prod_{j=1}^n X_j \to [0,\infty]$ be $\bigotimes_{j=1}^n \mathcal M_j$ measureable.
Note that $\bigotimes_{j=1}^n \mathcal{M}_j = \mathcal M_1 \otimes \bigotimes_{j=2}^n \mathcal M_j$, that $\mu_1 \times \mu_2 \times \cdots \times \mu_n = \mu_1 \times (\mu_2 \times \cdots \times \mu_n)$, and that $(\prod_{j=2}^n X_j, \bigotimes_{j=2}^n \mathcal M_j, \mu_2 \times \cdots \mu_n)$ is $\sigma$-finite (this is direct from proving the product of $\sigma$-finite measure spaces is $\sigma$-finite and induction).
Thus repeated application of the first application of Tonelli's theorem above (i.e. an induction) gives that:
$$\int f d(\mu_1 \times \cdots \times \mu_n) = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_1(x_1)\right) \cdots \right) d\mu_n(x_n).$$
Now we show inductively show that, under the assumptions of our proposition, that:
$$\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_1(x_1)\right) \cdots \right) d\mu_n(x_n) = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_{\sigma(1)}(x_{\sigma(1)})\right) \cdots \right) d\mu_{\sigma(n)}(x_{\sigma(n)})$$
for any permutation $\sigma \in S_n$.
It is clearly true for $n=1$.
Suppose it has been shown for $n$, then choose $\sigma \in S_{n+1}$. Then define $\tau \in S_n$ inductively by $\tau(1) = \sigma(1)$ if $\sigma(1) \ne n+1$ else $\sigma(2)$ and $\tau(j+1) = \sigma(\sigma^{-1}(\tau(j))+1)$ if $\sigma(\sigma^{-1}(\tau(j))+1) \ne n+1$ else $= \sigma(\sigma^{-1}(\tau(j))+2)$.
The result is that $\tau$ arranges $1,...,n$ in the same order as $\sigma$. Then applying the inductive hypothesis with $\tau$ to the interior integral for each $x_{n+1}$:
$$\int \left (\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_1(x_1)\right) \cdots \right) d\mu_n(x_n) \right) d\mu_{n+1}(x_{n+1}) = \int \left(\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_{\tau(1)}(x_{\tau(1)})\right) \cdots \right) d\mu_{\tau(n)}(x_{\tau(n)})\right) d\mu_{n+1}(x_{n+1}).$$
Then since $\tau$ put $1,...,n$ into the same order as $\sigma$, all that is left to get $1,...,n+1$ into the order induced by $\sigma$ is to insert $d\mu_{n+1}(x_{n+1})$ in the right spot, for which it suffices to show that two adjacent $d\mu_i(x_i)$ and $d\mu_j(x_j)$ can be commuted (then repeatedly commuting $d\mu_{n+1}(x_{n+1})$ left until it is in the right spot finishes the proof).
This we will now do. Claim:
$$\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_i \bigg) d\mu_j \cdots \right) d\mu_b(x_n) = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_j \bigg) d\mu_i \cdots \right) d\mu_b(x_n).$$
But this is just a direct application of Tonelli's theorem, since it suffices to show that:
$$\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_i \right) d\mu_j = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_j \right) d\mu_i,$$
and we have:
$$\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_i \right) d\mu_j = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \right) d\mu_j \times \mu_i = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_j \right) d\mu_i.$$
Putting it together completes the proof.
Note: Alternatively, instead of all that $\tau$ stuff, we can use the final claim to show that the set of permutations of the measures is a subgroup containing consecutive permutations: $(i, i+1)$ and then prove that $(i, i+1)$ generates $S_n$, which effectively what I did in the "$\tau$-section", though it might be a bit confusing.
Best Answer
For your first question:
We have that $$ \nu \times \mu (A):=\int \chi _A \,\mathrm d (\nu ,\mu )\tag1 $$ Now note that $A=\bigcup_{n\in \Bbb N }\{n\}\times [A]_n$, where $[A]_n:=\{y\in Y:(n,y)\in A \}$ and that $(\{n\}\times [A]_n)\cap (\{m\}\times [A]_m)=\emptyset $ if $n\neq m$, therefore $$ \nu \times \mu (A)=\sum_{n\in \Bbb N }\nu \times \mu (\{n\}\times [A]_n) =\sum_{n\in \Bbb N }\int \chi _{\{n\}\times [A]_n}\,\mathrm d (\nu,\mu )\tag2 $$ Now from the definition of product measure we knows that $$ \nu\times \mu (\{n\}\times [A]_n)=\mu (\{n\})\cdot \nu ([A]_n)=\nu([A]_n)\tag3 $$ Thus $$ \nu \times \mu (\{n\}\times [A]_n)=\iint \chi _{\{n\}\times [A]_n}\,\mathrm d \nu \,\mathrm d \mu =\iint \chi _{\{n\}\times [A]_n}\,\mathrm d \mu \,\mathrm d \nu\tag4 $$ where we used the fact that $\chi _{\{n\}\times [A]_n}=\chi _{\{n\}}\cdot \chi _{[A]_n}$. Then by the monotone convergence theorem we find that $$ \sum_{n\in \Bbb N }\iint \chi _{\{n\}\times [A]_n}\,\mathrm d \mu \,\mathrm d \nu=\iint \sum_{n\in \Bbb N }\chi _{\{n\}\times [A]_n}\,\mathrm d \mu \,\mathrm d \nu=\iint \chi_A \,\mathrm d \mu \,\mathrm d \nu\tag5 $$ that in view's of $\rm(2)$ and $\rm(4)$ shows that Tonelli's theorem holds.
For your second question:
In view of the above it seems that when $X$ is countable and the $\sigma$-algebra on $X$ is it power set then the product measure is unique.