No, for dimension $ >2 $ not every locally symmetric space has constant curvature. In fact it is not even the case that every symmetric space has constant curvature. For example take $ S^2 \times S^1 $, a model for one of the eight Thurston Geometries. The $ S^2 $ is round (constant curvature $ 1 $) while the $ S^1 $ is flat (constant curvature $ 0 $). The space $ S^2 \times S^1 $ is locally symmetric (indeed it is even symmetric since it is a product of symmetric spaces). But $ S^2 \times S^1 $ cannot have constant curvature because there is no way to but the same curvature on the $ S^2 $ piece and the $ S^1 $ piece.
Although not every locally symmetric space has constant curvature, it is the case that every constant curvature space is locally symmetric. Indeed every constant curvature space has a locally isometric universal cover by $ X=G/H $ for either $ X=S^n $ if it has constant positive curvature or $ X=E^n $ if it has $ 0 $ curvature or $ X=H^n $ if it has constant negative curvature. And thus every constant curvature space can be written as a locally symmetric space as
$$
\Gamma \backslash G/ H
$$
where $ \Gamma $ is a discrete subgroup of isometries acting freely on $ X=G/H $ (here $ G $ is the isometry group of $ X $, and $ X $ is the simply connected symmetric space of the appropriate curvature).
And yes the canonical way to pick $ \Gamma, G, H $ for a given locally symmetric space $ M $ is to take $ X $ to be the unique simply connected symmetric space covering $ M $ then $ G $ is the isometry group of $ X $ and $ H $ is the isotropy group of the action of $ G $ on $ X $ and $ \Gamma $ is the unique subgroup of isometries acting freely on $ X $ that commutes with the universal covering map.
I will write $S^n(r)$ for the round $n$-sphere of radius $r$. It is well known that the scalar curvature is a constant $\frac{n(n-1)}{r^2}$.
Now, for two positive real numbers $r,s$, consider the Riemannian product $S^2(r)\times S^2(s)$. This has scalar curvature $\frac{2}{r^2} + \frac{2}{s^2}$.
A relatively uninspired calculation shows that the scalar curvature has the constant value $2$ if and only if $r = \frac{s}{\sqrt{s^2-1}}$. (The value of $2$ was picked just to make this formula nice - nothing that follows depends critically on this choice.) In particular, there are infinitely many distinct pairs $(r,s)$ which give the same scalar curvature.
However, I claim that "most" of these pairs $(r,s)$ give different local isometry types. Specifically, the unordered pair $\{r,s\}$ controls the local isometry type. More precisely, I claim:
Proposition If $S^2(r)\times S^2(s)$ and $S^2(r')\times S^2(s')$ are locally isometric, then $(r,s)$ is a permutation of $(r',s')$.
Proof: We will assume without loss of generality that $r\geq s$ and that $r'\geq s'$.
Now, at every point of $S^2(r)\times S^2(s)$, the maximum sectional curvature of a two-plane is obtained by a $2$-plane tangent to the $S^2(s)$ factor, and it has sectional curvature $\frac{1}{s^2}$. Since local isometries preserve maximum curvature at a point, we must have $1/s^2 = 1/s'^2$, which easily implies $s = s'$. Moreover, the existence of a local isometry implies the scalar curvatures match, so $\frac{2}{r^2} + \frac{2}{s^2} = \frac{2}{r'^2}+\frac{2}{s'^2}$. As we already know $s = s'$, this simplifies to $\frac{2}{r^2} = \frac{2}{r'^2}$, which then implies $r= r'$. $\square$.
One can play a similar game with $S^m(r)\times S^n(s)$ as long as both $m,n\geq 2$. In particular, there are infinitely many such examples in each dimension $4$ or larger. However, one can not play this game with, say, $n=1$, basically because $S^1(s)$ is locally isometric to $S^1(s')$ even if $s\neq s'$. So the above approach won't work in dimension $3$.
However, there are three dimensional examples: the Berger spheres. A Berger sphere is obtained as follows: start with a round $3$-sphere $S^3(r)$ and the shorten the metric in the directions tangent to the Hopf fibration. If these directions are shortened by a factor of $t < 1$, then one can compute the resulting metric is homogeneous, and that, in an appropriate orthonormal basis of $T_p S^3$, that $K(X,Y) = K(X,Z) = r^2t^2$ while $K(Y,Z) = r^2(4-3t^2)$.
It follows that the scalar curvature is constant with value $r^2(8-2t^2)$. In particular, one can find infinitely many pairs $(r,t)$ with the same scalar curvature. However, note that for different choices of $t$, the ratio $\frac{\max{\text{sectional curvature}}}{\min{\text{sectional curvature}}} = \frac{4-3t^2}{t^2}$ is a local isometry invariant. In particular, $t$ is determined by the local isometry tyrpe, so there are infinitely many examples of different local isometry types but the same constant scalar curvature.
Best Answer
Suppose $\mathcal{M}$ is symmetric and $x \in \mathcal{M}$. Consider $s_x$ the symmetry at $x$, that is $$ s_x(\exp_x(tv)) = \exp_x(-tv) $$ It is an isometry by definition of symmetric space. Thus, $R$ is invariant under the action of $s_x$,. Now, one can show this general lemma: every $(r,s)$ tensor with $r+s$ odd that is invariant under $s_x$, is zero at $x$.
Apply this lemma to $\nabla R$, this will show that $(\nabla R)_x = 0$.