I thought of this question in Differential geometry class. Probably when topic is about self intersecting curves, professors draws (if) a curve coming from below, making a loop and going towards below on the board. Anyhow, I am not sure if we can form a curve that intersects itself at the same point infinitely many times.
Finitely many case is easy. I guess one way to construct is by choosing say vectors $v_1,v_2,...$ that are to be tangent vectors at times $t_1, t_2, ...$ for the curve $\gamma$ where for example we can let $t_n = 1 - \frac{1}{2^n}$ and curve 'speeds up' and comes back to the point of interest between $t_k - t_{k-1}$ but (at least) for this example since sequences of $t_n$ converge $1$, $\gamma(1) = \gamma(t_1)$ is restricted. Now, I can't exactly picture the behavior of this curve around $1$ which is the really only ambigious part, and I don't know if this defines a well-defined smooth map. Moreover, we can also find a subsequence $t_{n_k}$ such that $v_{n_k}$ converges to some $v$ so that we know the tangent vector at that point. Even more so, I can find all derivatives at $t = 1$ but still can't make sure if this is fine to do.
Note: Domain of the curve is $[0,1]$
This is a partial answer since it does not address the construction you propose in the second part of your question.
If the domain of the parametrized curve is, e.g., the open interval $I=(0,1)$, you can define the curve
\begin{align} \gamma : &\to \mathrm{R}^2\\ t &\mapsto \gamma(t)=\begin{pmatrix} \cos(-\ln(1-t))\\ \sin(-\ln(1-t)) \end{pmatrix}, \end{align} which intersects with itself infinitely (uncountably) many times at each point of its range.