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
Take a straight line embedding of the maximal planar graph. By maximality every face is a triangle. To show that it is $2$-connected, take any two faces $F_1$ and $F_2$, and any two points $p$ and $q$ in the interior of $F_1$ and $F_2$ respectively, and consider the line $(pq)$. For almost any choice of $p$ and $q$, this line does not intersect any vertex, and intersect each face on a (possibly empty) segment, except the outer face for which the intersection is the complement of a segment. Hence the intersected faces defines a circuit in the dual graph, containing $F_1$ and $F_2$ (as well as the outer face). As the connectivity between any two faces is at least 2, the graph is 2-connected, hence 2-edge-connected.
This also shows that your intuition is right: we can find two paths from every face to the outer face in the same way.