No, it is not always possible (even if $G$ is assumed to be discrete). For an example, let $N$ be the center of the discrete Heisenberg group: $$ N = \begin{bmatrix} 1 & 0 & \mathbb{Z} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \subset \begin{bmatrix} 1 & \mathbb{Z} & \mathbb{Z} \\ 0 & 1 & \mathbb{Z} \\ 0 & 0 & 1 \end{bmatrix} = G .$$ Let $\{a,b\}$ be a generating set of $G$, let $K = \{a,b,a^{-1},b^{-1}\}$, and let $z = [a,b] = a^{-1} b^{-1} a b \in N$. Then $$ z^{n^2} \sigma(1 \cdot N) = a^{-n} b^{-n} a^n b^n \, \sigma(1 \cdot N) = \sigma (a^{-n} b^{-n} a^n b^n \cdot N) \cdot \prod_{i=1}^{4n} \omega( s_i, x_i ) = \sigma (1 \cdot N) \cdot \prod_{i=1}^{4n} \omega( s_i, x_i ) ,$$ since $a^{-n} b^{-n} a^n b^n = z^{n^2} \in N$, where each $s_i$ is in $K$ and $x_i = s_{i-1} s_{i-1} \cdots s_1 N$. The exponent of $z$ on the left-hand side is a quadratic function of $n$, and the right-hand side has only linearly many terms in its product, so $\omega( s_i, x_i )$ cannot be a bounded element of $N$.

No, contrary to what you said, it is not always possible, even if $G$ is a semidirect product. Let us fix some $g \in N$, so $\sigma(gx) = \sigma(x)$ for all $x$. Then $\sigma(x)^{-1} g \sigma(x) = \omega(g,x)$ is a bounded function of $x$. It is easy to construct a counterexample to this, by arranging for the conjugacy class of $g$ to be unbounded. For example, let $G = \mathbb{R}^\times \ltimes \mathbb{R}$ be the $ax + b$ group with $N = \{(1,*)\}$ and take $g = (0,1)$. To eliminate this type of counterexample, you could replace $g$ on the right-hand side of your equation with $\sigma(g)$.

Here is a different type of counterexample (not a semidirect product). Let $N$ be the center of the discrete Heisenberg group: $$ N = \begin{bmatrix} 1 & 0 & \mathbb{Z} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \subset \begin{bmatrix} 1 & \mathbb{Z} & \mathbb{Z} \\ 0 & 1 & \mathbb{Z} \\ 0 & 0 & 1 \end{bmatrix} = G .$$ Let $\{a,b\}$ be a generating set of $G$, let $K = \{a,b,a^{-1},b^{-1}\}$, and let $z = [a,b] = a^{-1} b^{-1} a b \in N$. Then $$ z^{n^2} \sigma(1 \cdot N) = a^{-n} b^{-n} a^n b^n \, \sigma(1 \cdot N) = \sigma (a^{-n} b^{-n} a^n b^n \cdot N) \cdot \prod_{i=1}^{4n} \omega( s_i, x_i ) = \sigma (1 \cdot N) \cdot \prod_{i=1}^{4n} \omega( s_i, x_i ) ,$$ since $a^{-n} b^{-n} a^n b^n = z^{n^2} \in N$, where each $s_i$ is in $K$ and $x_i = s_{i-1} s_{i-1} \cdots s_1 N$. The exponent of $z$ on the left-hand side is a quadratic function of $n$, and the right-hand side has only linearly many terms in its product, so $\omega( s_i, x_i )$ cannot be a bounded element of $N$.

No (even if $G$ is assumed to be discrete). For an example, let $N$ be the center of the discrete Heisenberg group: $$ N = \begin{bmatrix} 1 & 0 & \mathbb{Z} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \subset \begin{bmatrix} 1 & \mathbb{Z} & \mathbb{Z} \\ 0 & 1 & \mathbb{Z} \\ 0 & 0 & 1 \end{bmatrix} = G .$$ Let $\{a,b\}$ be a generating set of $G$, let $K = \{a,b,a^{-1},b^{-1}\}$, and let $z = [a,b] = a^{-1} b^{-1} a b \in N$. Then $$ z^{n^2} \sigma(1 \cdot N) = a^{-n} b^{-n} a^n b^n \, \sigma(1 \cdot N) = \sigma (a^{-n} b^{-n} a^n b^n \cdot N) \cdot \prod_{i=1}^{4n} \omega( s_i, x_i ) = \sigma (1 \cdot N) \cdot \prod_{i=1}^{4n} \omega( s_i, x_i ) ,$$ since $a^{-n} b^{-n} a^n b^n = z^{n^2} \in N$, where each $s_i$ is in $K$ and $x_i = s_{i-1} s_{i-1} \cdots s_1 N$. The exponent of $z$ on the left-hand side is a quadratic function of $n$, and the right-hand side has only linearly many terms in its product, so $\omega( s_i, x_i )$ cannot be a bounded element of $N$.

No, it is not always possible (even if $G$ is assumed to be discrete). For an example, let $N$ be the center of the discrete Heisenberg group: $$ N = \begin{bmatrix} 1 & 0 & \mathbb{Z} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \subset \begin{bmatrix} 1 & \mathbb{Z} & \mathbb{Z} \\ 0 & 1 & \mathbb{Z} \\ 0 & 0 & 1 \end{bmatrix} = G .$$ Let $\{a,b\}$ be a generating set of $G$, let $K = \{a,b,a^{-1},b^{-1}\}$, and let $z = [a,b] = a^{-1} b^{-1} a b \in N$. Then $$ z^{n^2} \sigma(1 \cdot N) = a^{-n} b^{-n} a^n b^n \, \sigma(1 \cdot N) = \sigma (a^{-n} b^{-n} a^n b^n \cdot N) \cdot \prod_{i=1}^{4n} \omega( s_i, x_i ) = \sigma (1 \cdot N) \cdot \prod_{i=1}^{4n} \omega( s_i, x_i ) ,$$ since $a^{-n} b^{-n} a^n b^n = z^{n^2} \in N$, where each $s_i$ is in $K$ and $x_i = s_{i-1} s_{i-1} \cdots s_1 N$. The exponent of $z$ on the left-hand side is a quadratic function of $n$, and the right-hand side has only linearly many terms in its product, so $\omega( s_i, x_i )$ cannot be a bounded element of $N$.