Read a bit more carefully the definition that your book gives: "A directed graph may have multiple directed edges from a vertex to a second (possibly the same) vertex are called as directed multigraphs."
The key thing to notice here is that the multiple directed edges have the same origin and destination. Thus, in your first graph there is only one directed edge from vertex $c$ to vertex $d$ (and also only one directed edge from $d$ to $c$). So this graph is just a directed graph. On the other hand, in the second graph, there are two edges from $e$ to $d$, and two edges from $b$ to $c$. So this graph is a directed multigraph.
Yes; this result is one of the ones known as Petersen's theorem.
Assume $G$ is connected; if not, you can work with just one component of $G$ at a time. Then $G$ has an Eulerian tour $C$. By choosing one of the two directions around $C$, and then orienting each edge of $G$ according to which way $C$ goes along it, we get a directed graph. If $G$ was $2k$-regular, then every vertex of the directed version has in-degree and out-degree $k$.
We can encode a directed graph with $n$ vertices $v_1, v_2, \dots, v_n$ as a bipartite graph with $2n$ vertices $u_1, u_2, \dots, u_n$ and $w_1, w_2, \dots, w_n$: replace a directed edge $(v_i,v_j)$ by an edge $u_iw_j$ in the bipartite graph. If we do that to the directed graph we got, we get a $k$-regular bipartite graph.
In a $k$-regular bipartite graph, there is a perfect matching, by Hall's theorem: if $S$ is a set of vertices on one side of the graph and $N(S)$ their neighbors on the other side, then $$k|S| = \sum_{v \in S}\deg(v) = |E(S,N(S))| \le \sum_{w \in N(S)} \deg(w) = k|N(S)|$$ so $|S| \le |N(S)|$. (Key idea: summing the degrees out of $S$ counts all the edges between $S$ and $N(S)$, and summing the degrees out of $N(S)$ counts all those edges and possibly some others, if $N(S)$ has any edges to vertices not in $S$.)
The edges of the perfect matching form a subgraph of $G$ which visits each vertex exactly twice, so they give us the set of cycles we wanted. (In the multigraph case, a $2$-edge cycle is possible.)
Best Answer
The case where vertices are labeled would be easy to count. The tool that lets us go from labeled to unlabeled vertices is the Pólya enumeration theorem. (The Wikipedia article gives an example of using this to count undirected graphs with a fixed number of edges, and the directed case is only slightly different.)
I will forget about isolated vertices for now, and allow directed edges $vw$ and $wv$ to coexist. So let's start with the complete directed graph (with $n$ vertices and all $N := n(n-1)$ possible directed edges), and color the edges with two colors: a color that represents "there is an edge" (with weight $1$) and a color that represents "there is no edge" (with weight $0$). The generating function of the set of colors is $f(t) = 1+t$. The weight of a coloring is the total weight of the colors used, so it's the number of edges in our graph; therefore, we want to find the number of colorings with a fixed weight.
We take $G$ to be the group $S_n$ acting on the vertices of the complete directed graph. To apply the Pólya enumeration theorem as in the Wikipedia article, we need to compute $Z_G(t_1, t_2, \dots, t_N)$. So, for each permutation $\sigma$ of the vertices and each $k \in \{1, \dots, N\}$, we need to compute $c_k(\sigma)$, the number of $k$-cycles in the action of $\sigma$ on the set of edges.
For example, let's try this for $n=3$ vertices, with $N = 6$ possible edges. Here, there are $3$ types of permutations:
So we get $Z_G(t_1, t_2, t_3, t_4, t_5, t_6) = \frac16(t_1^6 + 3 t_2^3 + 2 t_3^2).$ We might as well make this $Z_G(t_1, t_2, t_3)$ since no longer cycles appear.
By the Pólya enumeration theorem, the generating function of directed graphs on $3$ vertices up to isomorphism is \begin{align} Z_G(1 + t, 1 + t^2, 1 + t^3) &= \frac16((1+t)^6 + 3(1+t^2)^3 + 2(1+t^3)^2) \\ &= 1 + t + 4 t^2 + 4 t^3 + 4 t^4 + t^5 + t^6 \end{align} so there are $1, 1, 4, 4, 4, 1, 1$ directed graphs on $3$ vertices with $0, 1, 2, 3, 4, 5, 6$ edges respectively. Here is a diagram of all these graphs to verify the count:
You also wanted to rule out isolated vertices, which this doesn't do, but that's easy to fix. Let's say that $a_{n,m}$ is the number of $n$-vertex $m$-edge digraphs without the isolated vertex condition, and $a_{n,m}^*$ is the number that have no isolated vertices. Then we have $$ a_{n,m} = a_{0,m}^* + a_{1,m}^* + a_{2,m}^* + \dots + a_{n,m}^* $$ because the number of $n$-vertex $m$-edge digraphs with $k$ isolated vertices is counted by $a_{n-k,m}^*$. From this, we can get $a_{n,m}^* = a_{n,m} - a_{n-1,m}$.