A $k$-degenerate graph needs only to have (at least) one vertex of degree at most $k$ in every subgraph, while the maximum degree is the largest degree of all vertices, which is completely unrelated. If you take the graph with $101$ vertices where one vertex is connected to all others but none of those vertices are connected, we have $\Delta(G)=100$, but every subgraph of $G$ contains a vertex of degree $1$, so $G$ is $1$-degenerate.
Quote from Wikipedia (at the examples section):
More strongly, the degeneracy of a graph equals its maximum vertex degree if and only if at least one of the connected components of the graph is regular of maximum degree. For all other graphs, the degeneracy is strictly less than the maximum degree.
In general, a "maximal property-$X$ graph" is a graph with property $X$ that has as many edges as possible in the weak sense that adding any edge not already present would destroy property $X$.
For example, a maximal acyclic graph is a tree, a maximal bipartite graph is a complete bipartite graph, and a maximal $k$-colorable graph is a complete $k$-partite graph.
So a maximal $k$-degenerate graph is a graph with the property that if you add any other edge to it, it stops being $k$-degenerate. You don't need to know what a $k$-monocore graph is to understand this definition.
It appears, however, that a $k$-monocore graph is a $k$-degenerate graph with minimum degree $k$. For graphs with at least $k+1$ vertices, we can prove that a maximal $k$-degenerate graph is a $k$-monocore graph:
First, let $G$ be a maximal $k$-degenerate graph and $v_1, v_2, \dots, v_n$ be a vertex ordering such that $v_i$ has at most $k$ edges to $v_{i+1}, \dots, v_n$. (Having such a vertex ordering is equivalent to being $k$-degenerate.) Then the vertices $v_{n-k}, v_{n-k-1}, \dots, v_n$ must form a clique (and therefore have degree $\ge k$), by maximality: any missing edges between them could be added without ruining the vertex ordering. For any other vertex $v_i$, if it had degree less than $k$, we could bring its degree up to $k$ by adding edges to some of the last $k+1$ vertices without ruining the ordering; therefore, by maximality, $G$ has minimum degree at least $k$. In fact, it has minimum degree exactly $k$ because the degree of $v_1$ is $k$.
I don't know what makes $k$-monocore graphs be "upper extremal".
Best Answer
Each edge of a subgraph must have two vertices. (These vertices need to be in the subgraph.) That is, a single vertex with one or more edges from it, but with the edges not connected to other vertices of the subgraph at their "other ends", is not considered a subgraph.