It doesn't matter that $G$ was chosen randomly. The choice of $G$ might matter if you asked for something more complicated about the distribution than the expected value.

The probability that a vertex $v$ is included is $M/N$.

Let the degree of $v$ be $h \ge k$ in $G$. The chance that precisely $k$ of its neighbors are included in $U$, conditioned on the inclusion of $v$, is 

$$\frac{{h \choose k}{N-h-1 \choose M-k-1}}{N-1 \choose M-1}. $$ 

So, the expected number of vertices of degree $k$ in $U$ is

$$ \sum_{h \ge k} d_h \frac{M}{N}\frac{{h \choose k}{N-h-1 \choose M-k-1}}{N-1 \choose M-1}.$$