Simple minded details about why a particular map is étale
Lemma 1 Let $A$ be a commutative ring in which no nonzero integer is a zero divisor (in particular, $A$ has characteristic $0$). Let $s \in A$ such that $s^2 = 1$. If $1-s$ is nilpotent, then $s = 1$.
Proof. Choose a positive integer $n$ such that $(1-s)^{2n+1} = 0$. By the binomial theorem, $$0 = \sum_{i=0}^{2n+1} \binom{2n+1}{i} (-1)^i s^i = \sum_{i=0}^n \binom{2n+1}{2i} - s \sum_{i=0}^n \binom{2n+1}{2i+1} = (1-s) \sum_{i=0}^n \binom{2n+1}{2i}.$$ Since $\sum_{i=0}^n \binom{2n+1}{2i}$ is not a zero divisor, $1-s = 0$, so $s = 1$. $\square$
Corollary If $k$ exists then it is unique.
Proof. Suppose there are two such maps, $k$ and $k'$. Since $kf = g = k'f$, we see that $k(x)^2 = k'(x)^2$ is a unit in $C$. In particular, $k(x)$ is a unit. Since $qk = h = qk'$, we see that $k(x) - k'(x)$ is nilpotent, so $1 - k'(x)/k(x)$ is nilpotent. By Lemma 1, $k'(x)/k(x) = 1$, i.e. $k(x) = k'(x)$. This shows that $k = k'$. $\square$
Existence of $k$ boils down to smoothness of the map $\operatorname{Spec} \mathbb{C}[t,t^{-1},x]/(x^2 - t) \to \operatorname{Spec} \mathbb{C}[t,t^{-1}]$, which we can unpack into straight algebra.
Lemma 2 If $N^2 = 0$ then there exists a map $k$ as desired.
Proof. Since $q$ is surjective, choose some $y \in C$ such that $q(y) = h(x)$. Now $q(y^2 - g(t)) = 0$, so $y^2 - g(t) \in N$. Since $q(2y) = h(2x)$ is a unit in $C/N$, $2y$ is a unit in $C$. So let $z = (g(t) - y^2)/(2y)$ (whence $z \in N$). Let $k_0 : \mathbb{C}[t,t^{-1},x] \to C$ be defined by sending $t \mapsto g(t)$ and $x \mapsto y+z$. Then $k_0(x^2 - t) = (y+z)^2 - g(t) = y^2 + (2y)z + z^2 - g(t) = y^2 - g(t) + (2y)z = 0$ since $N^2 = 0$. Thus, $k_0$ descends to give a map $k : \mathbb{C}[t,t^{-1},x]/(x^2 - t) \to C$ where $k(t) = g(t)$ and $k(x) = y + z$. It's easy to check that $kf = g$ and $qk = h$. $\square$
Now we can lift from $C/N$ to $C/N^2$, to $C/N^4$, ..., until we reach $C/N^m = C$. Formally, as an inductive proof:
Corollary There exists a map $k$ as desired.
Proof. Let $n$ be the smallest positive integer such that $N^n = 0$. We will use induction on $n$. If $n = 1$ then $q$ is an isomorphism and we are done. For the inductive step, suppose $n > 1$ and that the result holds for all smaller values of $n$. Factor $q$ as the composition of the quotients $C \to C/N^2 \to (C/N^2)/(N/N^2) \cong C/N$. Since $(N/N^2)^2 = 0$, Lemma 2 says that $h$ lifts up the projection $C/N^2 \to C/N$ to give a map $\ell : \mathbb{C}[t,t^{-1},x]/(x^2 - t) \to C/N^2$ making everything commute. By inductive hypothesis, $\ell$ lifts up to $k : \mathbb{C}[t,t^{-1},x]/(x^2 - t) \to C$. $\square$
Let me give a bare-hands answer.
Consider the Taylor expansion for $(1+u)^{1/2}$: $$ s(u) = \sum_{n\geq 0} \binom{1/2}{n} u^n.$$ This power series converges in a small disk around $1$ in $\mathbb C$, so at the level of formal power series with coefficients in $\mathbb Z[1/2]$ as well we have $s(u)^2 = 1+ u$. Thus, for any nilpotent element $v$ of a ring where $2$ is a unit, we have $s(v)^2 = 1+v$.
Now suppose we are in your situation. We are given $g(t) \in C$ a unit and some $y \in C$ such that $y^2 - g(t) \in N$. We wish to find a square root of $g(t)$ in $C$. But $ g(t)/y^2 - 1 \in N$, so $$s( (g(t)/y^2 - 1))^2 = 1 + (g(t)/y^2 - 1) = g(t)/y^2$$ and thus $$v = y s(g(t)/y^2 - 1)$$ is a square root of $g(t)$ in $C$. Uniqueness follows since if $v$ and $w$ are square roots, then $w^2/y^2 - 1 = g(t)/y^2 - 1 \in N$, so $w/y = s(1 + ((w/y)^2 - 1)) = s(1 + ((v/y)^2 - 1)) = v/y$ as we have an equality $s(1 + (u^2 -1 )) = u$ of formal functions.
This method seems mysterious on the face of it, relying on the Taylor expansion of the square root. I suspect it is more general.