The answer to your first question is a resounding no. An example (among many) is
given by $X=\mathrm{Spec} k$, $R=k[x]/(x^2)$ and $M=k$ considered as an
$R$-module through the $k$-algebra homomorphism given by $x\mapsto 0$.

As for the second question, the reason is that in general there are correction
terms to this formula coming from the failure of $M$ being locally
projective. In fact there is a long exact sequence
$$
0\rightarrow H^0(X,\mathcal{E}\mathrm{xt}^1_R(M,M))\rightarrow
\mathrm{Ext}^1_R(M,M)\rightarrow H^1(X,\mathcal{H}\mathrm{om}_R(M,M))
\rightarrow H^0(X,\mathcal{E}\mathrm{xt}^2_R(M,M)),
$$
where the $\mathcal{E}\mathrm{xt}^i_R(M,M)$ are the sheaves of Ext-classes
(their stacks at $x$ are the $\mathrm{Ext}^i_{R_x}(M_x,M_x)$). Hence, you need
something like (possibly  something a little bit weaker) the vanishing
$\mathrm{Ext}^i_{R_x}(M_x,M_x)$ for $i=1,2$ which in turn are implied by (though not
implying) the local $R$-projectivity of $M$.