Here's an elementary observation that gives a relation between the three sets in both cases.

Note that $M_1\# M_2$ admits degree one maps $\alpha : M_1\# M_2 \to M_1$ and $\beta : M_1\#M_2 \to M_2$ (just crush the other summand).

If $f : M_1 \to N$ has degree $d$, then $f\circ\alpha : M_1\# M_2 \to N$ also has degree $d$. Therefore $D(M_1, N) \subseteq D(M_1\# M_2, N)$ and likewise $D(M_2, N) \subseteq D(M_1\# M_2, N)$ so

$$D(M_1, N)\cup D(M_2, N) \subseteq D(M_1\#M_2, N).$$

If $g : N \to M_1\# M_2$ has degree $d$, then $\alpha\circ g : N \to M_1$ also has degree $d$. Therefore $D(N, M_1\# M_2) \subseteq D(N, M_1)$ and likewise $D(N, M_1\# M_2) \subseteq D(N, M_2)$ so

$$D(N, M_1\# M_2) \subseteq D(N, M_1)\cap D(N, M_2).$$

In general, the inclusions are strict as can be seen by using Jens Reinhold's answer: take $M_1 = M_2 = N = \Sigma_2$ for example.

There are also counterexamples in higher dimensions.

Let $d \geq 3$. If $M$ and $N$ are closed, connected, oriented manifolds of dimension $d$, then $\|M\# N\| = \|M\| + \|N\|$; note, this is not the case in dimension two (the torus has Gromov norm zero, but higher genus surfaces do not).

Suppose $M$ has dimension $d$ and $\|M\| \neq 0$. If $f : M \to M\# M$, then

$$\|M\| \geq |\deg f|\|M\# M\| = 2|\deg f|\|M\|$$

so $f$ has degree zero, i.e. $D(M, M\#M) = \{0\}$. Note however that $1 \in D(M\# M, M\#M)$ so

$$D(M, M\#M)\cup D(M, M\# M) \subsetneq D(M\#M, M\#M).$$

Also $1 \in D(M, M)$ so

$$D(M, M\#M) \subsetneq D(M, M)\cap D(M, M).$$

Gromov proved that the Gromov norm of a closed, oriented, hyperbolic $n$-manifold is a non-zero multiple (depending only on $n$) of its volume. In particular, such manifolds have non-zero Gromov norm so they can be used to construct counterexamples in any dimension greater than or equal to $2$.

These counterexamples seem to suggest that the answer to both questions of your questions is no.