Let $G$ be a $4$-regular maximal planar graph. By Handshaking lemma
$$2m = \sum_{v \in V(G)} \deg v = 4n.$$
So $2n = m = 3n - 6$, therefore $n = 6$. Then $\overline{G}$ is $1$-regular graph on $6$ vertices, that is $3K_2$. Then $G \cong \overline{3K_2}$ is the only $4$-regular maximal planar graph.
Given a planar graph, embedded in the plane, you want to add a number of edges to get a maximal planar graph, and you want to be sure that resulting maximal planar graph won't have vertices with degree $=2$.
From this Wikipedia page:
A simple graph is called maximal planar if it is planar but adding any edge (on the given vertex set) would destroy that property...
The process of adding edges is straightforward - you need to find all faces, bounded by more than $3$ edges (including the external face), and triangulate these faces by adding new edges.
Let's consider the case, which looks problematic to you - a vertex $u$ with degree $2$, for which it's impossible to add a new edge, incident to the $u$. You have two vertices $v$ and $w$, adjacent to the vertex $u$, and two faces, partially bounded by edges $\{v,u\}$ and $\{u,w\}$. If a new edge, incident to the vertex $u$, can't be added, then both these faces should have an edge $\{v,w\}$, which prevents this addition. So, the graph contains two parallel edges $\{v,w\}$, which is not possible, because we consider simple graphs only.
Best Answer
Yes, this is essentially correct. There is one slight niggle: you have assumed the number of edges for a maximal planar graph on $k$ vertices is $3k-6$, but that only holds for $k\geq 3$. Since you apply this with $k=n-1$, your argument assumes $n\geq 4$ and you need to check the (easy) case $n=3$ separately.