Here is an elementary proof that the answer is positive, at least if you relax the condition that the extension be $C^2$.
Lemma. Given $x_0\in C_M$, there exists a linear form $\lambda$ such that $f(x_0)+\lambda(x-x_0)\ge0$ over the graphs of both $f$ and $g$.
To see this, consider the tangent space $\{(x,f(x_0)+df_{x_0}(x-x_0))\,|\,x\in C_M\}$ to the graph of $f$. It is located below this graph (touching at $x_0$). If it meats the graph of $G$ at some point $(y,p)\in C_N$, then $y\in M\cap N$ and therefore $f(y)=g(y)$. But then $f(y)=p=f(x_0)+df_{x_0}(y-x_0)$, which implies $y=x_0$. There are therefore two cases. Either $x_0\in C_M\cap C_N$, and because $dg_{x_0}=df_{x_0}$ over $M\cap N$, we may choose $\lambda$ an compatible extension of $dg_{x_0}$ and $df_{x_0}$ to $R^d$. Or $x_0\not\in C_N$, and then you can choose $\lambda$ by applying Hahn-Banach.
Let me now remark that if $x_0$ and $\lambda$ are as in the lemma, then by assumption, the function $$\phi_{x_0,\lambda}(x):=\frac\delta2|x-x_0|^2+\lambda(x-x_0)+f(x_0)$$ satisfies $\phi_{x_0,\lambda}|_{C_M}\le f$ and $\phi_{x_0,\lambda}|_{C_N}\le g$.
Likewise, we consider those pairs $(y_0,\mu)$ such that the functions $$\psi_{y_0,\mu}(x):=\frac\delta2|x-y_0|^2+\mu(x-y_0)+g(y_0)$$ satisfy $\psi_{y_0,\mu}|_{C_M}\le f$ and $\psi_{y_0,\mu}|_{C_N}\le g$.
Finally, we form the function $$S(x)=\max\left\{\max_{x_0,\lambda}\phi_{x_0,\lambda}(x),\max_{y_0,\mu}\psi_{y_0,\mu}(x)\right\}.$$ This is a convex function, which coincides with $f$ over $C_M$ and with $g$ over $C_N$. In addition, it satisfies $D^2S\ge\delta I_d$ in the sense of distributions. In other words, for every $z_0$, there exists a linear form $\zeta$ such that $$S(x)\le S(z_0)+\zeta(x-z_0)+\frac\delta2|x-z_0|^2,\qquad\forall\,x.$$
Notice that $S$ is the largest function satisfying all the queries. Except the $C^2$ regularity of course ; exactly as an ordinary convex enveloppe fails to be $C^1$.