The mean value theorem for a real valued $C^1$ function gives $|\phi(x_1)-\phi(x_0)| \leq \max_{\xi \in [x_0,x_1]} |\phi'(\xi)|\|x_1-x_0\|$, so if $|\phi'(x)|$ is bounded by $L$ on some convex set, then $|\phi(x_1)-\phi(x_0)| \leq L \|x_1-x_0\|$ on this set and hence is uniformly continuous (given $\epsilon>0$, choose $\delta = \frac{\epsilon}{L}$).
If $\phi$ is a vector-valued function (ie, $\phi(x) = (\phi_1(x),...,\phi_n(x))^T$), and each of the derivatives $\phi_i'$ is bounded by some $L$ on some convex set, then $\|\phi(x_1)-\phi(x_0)\| \le \sqrt{\sum_i(\phi_i(x_1)-\phi_i(x_0))^2} \leq \sqrt{\sum_i L^2 \|x_1-x_0\|^2} = \sqrt{n} L \|x_1-x_0\|$ on this set, and hence is uniformly continuous.
For 1), it is easy to see that each element of $DF((x,y))$ is bounded by $2$, hence $f$ is uniformly continuous on $\mathbb{R}^2$.
For 2), $Df((x,y)) = \begin{bmatrix} \frac{y^2-2xy-x^2}{(x+y)^2} & \frac{y^2+2xy-x^2}{(x+y)^2} \end{bmatrix}$. For $(x,y) \in M$, $|y^2-2xy-x^2| \leq (x+y)^2$, and $|y^2+2xy-x^2| \leq (x+y)^2$, hence each entry is bounded by $1$. Hence $f$ is uniformly continuous on $M$.