The curve $C$ is defined by a level-set function $f(x,y)$ as
$$
C := \big\{(x,y)\in \mathbb{R}^2\, \big|\ \ f(x,y) = a \big\}.
$$
Note that the function $f:\mathbb{R}^2 \to \mathbb{R}$ is defined on a two-dimensional domain; therefore you cannot use this function for describing anything in $\mathbb{R}^3\!$. However, you were asked to define a surface using the same equation. So, technically what you do is you define an extension of $f$ in $\mathbb{R}^3\!$. Very often people abuse notations and use the same symbol for the extension and for original function. For the sake of clarity, let us avoid this common practice here.
Define the extension $F: \mathbb{R}^3 \to \mathbb{R}$ of $f$ into $\mathbb{R}^3$. This extension can be defined as $ F(x,y,z) : = f(x,y)$. Then, define generalized cylinder as the following set:
$$
\mathfrak{C} := \big\{(x,y,z)\in \mathbb{R}^3\, \big|\ \ F(x,y,z) = a \big\}.
$$
Using definition of $F$, we can write
$$
\mathfrak{C} := \big\{(x,y,z)\in \mathbb{R}^3\, \big|\ \ f(x,y) = a \big\} .
$$
But then
$$
\mathfrak{C} = \big\{(x,y)\in \mathbb{R}^2\, \big|\ \ f(x,y) = a \big\}
\times \big\{z\in \mathbb{R} \big\} = C \times P,\tag{*}\label{0}
$$
where $P := \{z\in \mathbb{R} \} $ and $\times$ stands for Cartesian product.
Observe that
- $C\in\mathbb{R}^2 $ is a plane curve, i.e. a $1$-dimentional smooth manifold embedded into $\mathbb{R}^2\!$,
- $P \sim \mathbb{R}$ is a straight line, which is also a $1$-dimentional smooth manifold, we have
- $C \not\subseteq P$ and $C \not\supseteq P$ because $P$ is orthogonal to the plane of $C$.
In this way, our cylinder $\mathfrak{C}$ is defined as a Cartesian product of smooth manifolds of dimensionality $1$. Therefore $\mathfrak{C}$ is itself a smooth manifold of dimension $2$ embedded into $\mathbb{R}^3\!$!
Now we only have to derive parametrization of $\mathfrak{C}$.
First, recall that any plane curve $\gamma$ can parametrized in $\mathbb{R}^2$ by a smooth map
$$
\gamma : I \to \mathbb{R}^2,
$$
where $I\in\mathbb{R}$ is some interval ( a non-empty connected subset of $\mathbb{R}$) and $\gamma(t) = \big[\,x(t),\, y(t)\, \big]'$.
Since we need the parametrization in $\mathbb{R}^3$, we can use extension of $\gamma$ and define a new parametrization
$$
\beta(t) =
\begin{bmatrix}
x(t)\\ y(t) \\ 0
\end{bmatrix}
: I \to \mathbb{R}^3, \qquad \text{ where } \quad
\gamma(t) =
\begin{bmatrix}
x(t)\\ y(t)
\end{bmatrix}\tag{1}\label{1}
$$
Second, denote $\vec{n} = [0,0,1]$ — the unit normal vector for domain plane of $C$.
Then $P$ can be parametrized in $\mathbb{R}^3$ by the map $\mathbf{v}: \mathbb{R} \to \mathbb{R}^3$
$$\tag{2}\label{2}
\alpha(s) = s\,\vec{\mathbf{n}} =
\begin{bmatrix}
0 \\ 0 \\ s
\end{bmatrix},
\quad s \in \mathbb{R}
$$
Finally, combining parameterizations $\eqref{1}$ and $\eqref{2}$ in the Cartesian product expression $\eqref{0}$ for $\mathfrak{C}$, we get the following parametrization for generalized cylinders:
$$
\mathbf{r}(t,s) = \beta(t) + \alpha(s) = \beta(t) + s\,\vec{\mathbf{n}},
$$
where $t\in I\subset\mathbb{R}$, $s\in\mathbb{R}$, $\beta(t)= \big[\,x(t),\, y(t),\, 0\,\big]'$, and $\vec{\mathbf{n}} = [0,0,1]'$
