Can we show that $\mathbb R^{\mathbb N}\ni x\mapsto\bigotimes_{n\in\mathbb N}\mathcal N_{x,\:\sigma^2}$ is a Markov kernel? Let $\sigma>0$ and $\mathcal N_{x,\:\sigma^2}$ denote the normal distribution with mean $x\in\mathbb R$ and variance $\sigma^2$. From the Ionescu-Tulcea theorem, we know that $$\kappa(x,\;\cdot\;):=\bigotimes_{n\in\mathbb N}\mathcal N_{x_n,\:\sigma^2}$$ is a well-defined probability measure on $\mathcal B(\mathbb R)^{\otimes\mathbb N}$ for all $x\in\mathbb R^{\mathbb N}$.

Are we able to show that $$\mathbb R^{\mathbb N}\ni x\mapsto\kappa(x,B)$$ is Borel measurable for all $B\in\mathcal B(\mathbb R)^{\otimes\mathbb N}$?

If the claim is true, $\kappa$ would be a Markov kernel on $\left(\mathbb R^{\mathbb N},\mathcal B(\mathbb R)^{\otimes\mathbb N}\right)$. Are we able to infer that then the restriction of $\kappa$ to $[0,1]^{\mathbb N}\times\mathcal B([0,1])^{\otimes\mathbb N}$ is a Markov kernel on $\left([0,1]^{\mathbb N},\mathcal B([0,1])^{\otimes\mathbb N}\right)$?
 A: First of all, I suppose you mean $\kappa$ to be defined as
$$\kappa(x,\;\cdot\;):=\bigotimes_{n\in\mathbb N}\mathcal N_{x_n,\:\sigma^2}$$
with $x_n$ on the right side instead of $x$, where $x = (x_1, x_2, \dots)$.  As originally written it didn't make sense.
Defined thus, $\kappa$ is indeed a Markov kernel.  As you note, we need to prove that $\kappa(\cdot,B)$ is measurable for every Borel set $B\in\mathcal B(\mathbb R)^{\otimes\mathbb N}$. One way is to use the Dynkin $\pi$-$\lambda$ lemma.  Let $\mathcal{L}$ be the collection of all sets $B\in\mathcal B(\mathbb R)^{\otimes\mathbb N}$ such that $\kappa(\cdot,B)$ is Borel measurable.  You may easily show that $\mathcal{L}$ is a $\lambda$-system:


*

*When $B = \mathbb{R}^{\mathbb{N}}$, we have $\kappa(x,B)=1$ for every $x$, and the constant function $1$ is measurable.  So  $\mathbb{R}^{\mathbb{N}} \in \mathcal{L}$.

*If $B \in \mathcal{L}$, then $\kappa(x,B^c) = 1-\kappa(x,B)$ for every $x$, because $\kappa(x,\cdot)$ is a probability measure.  So $\kappa(\cdot, B^c) = 1-\kappa(\cdot, B)$ is measurable because $\kappa(\cdot, B)$ was.

*If $B_1, B_2, \dots \in \mathcal{L}$ are disjoint, and $B = \bigcup_k B_k$, then we have $\kappa(\cdot, B) = \sum_{k=1}^\infty \kappa(\cdot, B_k)$ which is measurable since it is an infinite sum of measurable functions.
Let $\mathcal{P}$ be the collection of all "rectangles" of the form $B = B_1 \times B_2 \times \dots \times B_m \times \mathbb{R} \times \mathbb{R} \times \dots$, where $m$ is an integer and $B_1, \dots, B_m \in \mathcal{B}(\mathbb{R})$.  Clearly $\sigma(\mathcal{P}) = \mathcal B(\mathbb R)^{\otimes\mathbb N}$.  So it remains to show that $\mathcal{P} \subset \mathcal{L}$.  But for such $B$ we have $\kappa(x,B) = \frac{1}{(2 \pi \sigma^2)^{m/2}} \prod_{k=1}^m \int_{B_k} \exp(-(x_k-y_k)^2/2\sigma^2)\,dy_k$ and it is easy to check this is actually a continuous function of $x$.
You can also construct a proof with the monotone class theorem if you like it better.  But either way, with some practice this kind of argument should be completely routine.
The Markov chain on $\mathbb{R}^{\mathbb{N}}$ whose transition kernel is $\kappa$ is easy to describe: the coordinates perform independent random walks, each of which has iid $\mathcal{N}_{0, \sigma^2}$ increments.
I am not sure I understand exactly what you mean by "restriction", but note for example that $\kappa(x, [0,1]^{\mathbb{N}}) = 0$ for all $x$.  (If you sample countably many normal random variables all with the same variance, there is zero probability that they will all come out between 0 and 1.)  So I don't see how to turn this into a Markov kernel on $[0,1]^{\mathbb{N}}$ in any sensible way.
