A maximal matching can have less than $n/2$ edges. Imagine a $C_6$ with the vertices named in order $a,b,c,d,e,f$. Then, the $ab$ and $ed$ edges form a maximal matching, and here it's $n/3$ edges.
So what's the most stupid way of choosing your matching ?
Let $X$ denote the set of vertices of $C_n$ that are incident to an edge of some matching $M$.
Then, $|M| = |X| / 2$. So how can you minimize $|X|$ while making $M$ maximal ?
You should be able to do the rest by observing that if $M$ is maximal, then for two adjacent vertices $v_1, v_2$, at least one has to be in $X$, because otherwise you can add the $v_1v_2$ edge in $M$.
We can prove the following theorem, which tells us considerably more about such graphs:
Theorem. If $\kappa(G) > \alpha'(G)$, then $\alpha'(G) = \lfloor \frac n2 \rfloor$: a maximum matching leaves at most one vertex of $G$ unmatched.
Proof. Suppose $\alpha'(G) = \frac{n-k}{2}$, so that a maximum matching has $k$ unmatched vertices, and that $k \ge 2$. Then by the Tutte–Berge formula there is a set of vertices $U$ such that $G-U$ has $|U|+k$ odd components. This is, in particular, multiple odd components: so $\kappa(G) \le |U|$, because $G-U$ is not connected. Also, $G-U$ must have at least $|U|+k$ vertices, so $n-|U| \ge |U|+k$, or $|U| \le \frac{n-k}{2}$. In summary, $$\kappa(G) \le |U| \le \frac{n-k}{2} = \alpha'(G).$$ Therefore if $\kappa(G) > \alpha'(G)$, then $k \le 1$ and so $\alpha'(G) \ge \frac{n-1}{2}$.
In fact, we don't need as strong an inequality as $\kappa(G) > \alpha'(G)$: if $\kappa'(G) > \alpha'(G)$, or even if $\delta(G) > \alpha'(G)$, then we can already conclude that $\alpha'(G) = \lfloor \frac n2 \rfloor$.
For this, we use the Tutte–Berge formula more carefully. As before, suppose $\alpha'(G) = \frac{n-k}{2}$ with $k\ge 2$, and take any set of vertices $U$ such that $G-U$ has $|U|+k$ odd components. Now, additionally, let the smallest odd component have $2j+1$ vertices.
Then $$\alpha'(G) = \frac{n-k}{2} \ge \frac{|U|+(2j+1)(|U|+k)-k}{2} = (j+1)|U| + jk$$ whereas $\delta(G) \le |U|+2j$: this is the maximum possible degree of a vertex in the smallest odd component. The inequality $\delta(G) > \alpha'(G)$ rearranges to $j(|U|+k-2) < 0$, contradicting our initial assumption that $k \ge 2$.
This is not a characterization, but I think that there will be no good description beyond this. For example, if you take a random $n$-vertex graph with edge probability $\frac12 + \epsilon$, then with high probability $\alpha'(G) = \lfloor \frac n2 \rfloor$ and $\kappa(G) = \kappa'(G) = \delta(G) \approx (\frac12 + \epsilon)n$. This gives lots and lots of boring examples.
Best Answer
You know, you're right. The post you linked to has too short an answer. Moreover, I don't see any proof of this statement there even for the case where $k<|V|/2$.
Here is what seems to me to be more plausible reasoning. Let $k<|V|/2$. If $k=0$, then there is nothing to prove. Let $k\geq1$.
Let's organize such a process. In the first step, choose an arbitrary vertex $x_1$. It has $k$ neighbors. Take its arbitrary neighbor $y_1$. We obtain an edge $x_1y_1$.
Let us already choose edges $x_1y_1,\ldots,x_sy_s$ where $x_1,y_1,\ldots,x_s,y_s$ are pairwise different and $s<k$. Let $X=\{x_1,\ldots,x_s\}$, $Y=\{y_1,\ldots,y_s\}$, and $Z=X\cup Y$.
If there is at least one edge outside $Z$, we can increase our matching. Let each vertex outside $Z$ be adjacent only to vertices from $Z$. Since $2s<2k<|V|$ there are at least two vertices $x,y\notin Z$.
If there exists $i\in[1,s]$ such that $ux_i,vy_i\in E(G)$, then we add these two edges to our matching and remove edge $x_iy_i$ from it and thus increase our matching.
Let us prove that such $i$ necessarily exists. Let $I_u\subset[1,s]$ and $J_u\subset[1,s]$ be chosen such that $ux_i\in E(G)$ and $uy_j\in E(G)$ for every $i\in I_u$ and every $j\in J_u$. Since $\operatorname{deg}(u)\geq k$, then $|I_u\cup J_u|\geq k>s$. Similarly determine $I_v$ and $J_v$. If $I_u\cap J_v\neq\emptyset$ or $I_v\cap J_u\neq\emptyset$, then the previous observation works. If $I_u\cap J_v=\emptyset$ and $I_v\cap J_u=\emptyset$, and for example $|I_u|>s/2$, then $|J_v|>s/2$, then $I_u\cap J_v\neq\emptyset$. Contradiction.
The case of $k\geq|V|/2$ is completely similar to the considered one. After all, in this case we have to build a matching of cardinality $l=|V|/2$.