Number of partitions for a complete graph, each with bounded degree I will assume $n$ to be large enough and calculate with give or take 1, that won't matter too much.
Given a complete graph $K_n = (V, E) = ([n], {[n] \choose 2})$ with $[n] = \{0, 1, ..., n-1\}$ and $ {[n] \choose 2}$ all subsets of $[n]$ of cardinality 2, I'm looking for a tight upper bound on $k = k(d)$ for the following:
\begin{align}
E = \bigcup_{1 \leq i \leq k} S_i, \; \text{with} \;\; \Delta((V, S_i)) \leq d
\end{align}
So I'm looking to partition $E$ into subsets such that in each induced subgraph, all vertices have degree smaller or equal to some $d \in \mathbb{N}^{+}$.
Clearly, for $d_1 \geq d_2$, we have $k(d_1) \leq k(d_2)$ since a solution with at most $d_2$ edges per node also has at most $d_1$ edges. Furthermore, $k(n-1) = 1$ as the complete $E$ can serve as a solution. For $d = 2$, using the group $([n], 0, f), f = \lambda x,y.(x+y)\%n$ of addition modulo $n$ gives a tight bound: for any fixed $m \in [n], f(\_, m)$ is bijective (and $f$ is commutative), so the subset $S_i$ contains an edge $\{a, b\}$ whenever $f(a, i) = b$.
For example, for $n = 6$, the above gives the cycles $0 \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 5 \rightarrow 0$ for $i = 1$, $0 \rightarrow 2 \rightarrow 4 \rightarrow 0$ and $1 \rightarrow 3 \rightarrow 5 \rightarrow1$ for $i = 2$ and $0 \rightarrow 3 \rightarrow 0$, $1 \rightarrow 4 \rightarrow 1$ and $2 \rightarrow 5 \rightarrow 2$ for $i = 3$, which translated into (un-)directed edges fullfill the criterion above. That's also tight since every node has degree 2.
This method for $d = 2$ gives $\frac{n}{2}$ subsets (the "upper" half ($m=4,5$ in the example) can be discarded in the undirected case), the union of which gives $E$ by totality of $f(\_, m)$.
By induction, for $d$ any even number, the number of subsets should be $\frac{n}{d}$ as one can just take the union of multiple of the subsets for $d=2$.
Question 1: How about the case of $d=1$? In the case that the group decomposes into cycles of even length only, $k = n$, e.g. take the solution for $d=2$ and split each subset $S_i$ into two. However, this does not work when there are "many" cycles of odd length in the decomposition. The only bound I could come up with was $k = n + o(n)$ with $o(n)$ the number of odd-length-producing-values (my algebra is not even strong enough to know what $o(n)$ is, maybe the number of odd prime factors of $n$, counting multiplicity?).
Question 1.1: Given a tight bound on $d=1$, how does that give a bound on $d > 2$ and odd? E.g. if $k=k(1) = n + o(n)$ were tight, is $k(3) = \frac{n}{3} + o(n)$?
Question 2: Given that I were to look at the directed case and I were to require the sum of in- and out-degree to be $\;\leq d$, would I be correct to assume the bound on $k$ is plainly twice as high as in the undirected case?
 A: When $d=1$, we are looking for a partition of $E(K_n)$ into matchings. A matching in an $n$-vertex graph can contain at most $\lfloor \frac n2\rfloor$ edges. When $n$ is even, this is $\frac n2$, and so we can hope to split the $\binom n2$ edges into $n-1$ matchings of $\frac n2$ edges each. When $n$ is odd, $\lfloor \frac n2 \rfloor = \frac{n-1}{2}$, and so at best we can hope to split the $\binom n2$ edges into $n$ matchings of $\frac{n-1}{2}$ edges each.
These are both achievable. A construction due to Soifer does the following:

*

*When $n$ is odd, put all $n$ vertices at the corners of a regular polygon. For each side of the polygon, take all the edges parallel to that side, and let that be one of the matchings (which saturates all but $1$ vertex: the vertex opposite the side we picked).

*When $n$ is even, do the above for $n-1$ of the vertices, and put the $n^{\text{th}}$ vertex at the center. In each matching found for the odd case, join the center vertex to the vertex unsaturated by the matching.

Wikipedia has an illustration of this under https://en.wikipedia.org/wiki/Edge_coloring#Examples.

What's best possible, for general $d$? Well, an $n$-vertex graph with maximum degree $d$ has at most $\frac{nd}{2}$ edges. We have $\frac{n(n-1)}{2}$ edges total, so we need at least $\frac{n(n-1)/2}{n d/2} = \frac{n-1}{d}$ parts in the partition. Therefore having $\lceil \frac{n-1}{d}\rceil$ parts is optimal whenever it's possible.
For any $d$, and when $n$ is even, we have a partition into $k = \lceil \frac{n-1}{d}\rceil$ edge sets and reach this lower bound, just by taking our $\frac n2$-edge matchings $d$ at a time until we run out. When $n$ is odd, this only gives us a partition into $k = \lceil \frac nd \rceil$ edge sets.
When $n$ is odd, we can do better in the $d=2$ case. Walecki's construction, a modification of the polygon strategy above, splits the edges of $K_n$ when $n$ is odd into $\frac{n-1}{2}$ cycles of length $n$. This achieves the bound for $d=2$, and we can achieve the same bound for any even $d$ as well, by taking the Hamiltonian cycles $\frac d2$ at a time until we run out.
Finally, when $n$ and $d$ are both odd, having $\lceil \frac{n}{d}\rceil$ parts is best possible. To see this, note that $\lceil \frac{n}{d}\rceil = \lceil \frac{n-1}{d}\rceil$ (achieving the optimal value) unless $\frac{n-1}{d}$ is an integer. But in that case, in order to achieve exactly $\frac{n-1}{d}$ parts, we need to split the $\binom{n(n-1)}{2}$ edges into $\frac{n-1}{d}$ groups of exactly $\frac{nd}{2}$ edges, and that's plainly impossible: $\frac{nd}{2}$ is not an integer. So $\frac{n-1}{d}$ parts is impossible, and we can't improve on $\lceil \frac{n}{d}\rceil$.
