We consider the Riemann tensor first. A crucial observation is that if we parallel
transport a vector $u$ at $p$ to $q$ along two different pathes $vw$ and $wv$, the resulting
vectors at $q$ are different in general (following figure). If, however, we parallel transport
a vector in a Euclidean space, where the parallel transport is defined in our
usual sense, the resulting vector does not depend on the path along which it
has been parallel transported. We expect that this non-integrability of parallel
transport characterizes the intrinsic notion of curvature, which does not depend
on the special coordinates chosen.
It is useful to say that in this sense visualization of the first Bianchi identity is very easy:
We can give a quantitative geometric interpretation to the sectional curvature
tensor in any dimension. Let M be a Riemannian n-manifold and p ∈ M. If $\Pi$ is any $2$-dimensional subspace of $T_pM$, and $V \subset T_pM$ is any neighborhood
of zero on which $\exp_p$ is a diffeomorphism, then $S_\Pi := \exp_p(\Pi \cap V)$ is a
$2$-dimensional submanifold of $M$ containing $p$ (following figure), called the plane
section determined by $\Pi$. Note that $S_\Pi$ is just the set swept out by geodesics
whose initial tangent vectors lie in $\Pi$. We define the sectional curvature of $M$ associated with $\Pi$, denoted $K(\Pi)$,
to be the Gaussian curvature of the surface $S_\Pi$ at $p$ with the induced metric. If $(X, Y)$ is any basis for $\Pi$, we also use the notation $K(X, Y)$ for $K(\Pi)$.
Proposition: If $(X, Y)$ is any basis for a $2$-plane $\Pi \subset T_pM$, then $$K(X,Y)=\frac{Rm(X,Y,Y,X)}{|X|^2 |Y|^2 -\langle X,Y \rangle ^2}$$
We can also give a geometric interpretation for the Ricci and scalar
curvatures. Given any unit vector $V \in T_pM$, choose an orthonormal basis
$\{E_i\}$ for $T_pM$ such that $E_1 = V$ . Then $Rc(V, V )$ is given by
$$Rc(V,V)=R_{11}=R_{k11}^k=\sum_{k=1}^{n} Rm(E_k,E_1,E_1,E_k)=\sum_{k=2}^{n}K(E_1,E_k)$$
Therefore the Ricci tensor has the following interpretation: For any unit
vector $V \in T_pM$, $Rc(V, V )$ is the sum of the sectional curvatures of planes
spanned by $V$ and other elements of an orthonormal basis. Since $Rc$ is symmetric
and bilinear, it is completely determined by its values of the form
$Rc(V, V )$ for unit vectors $V$ .
Similarly, the scalar curvature is
$$S=R_j^j=\sum_{j=1}^n Rc(E_j,E_j)=\sum_{j,k=1}^{n}Rm(E_k,E_j,E_j,E_k)=\sum_{j\ne k}K(E_j,E_k)$$
Therefore the scalar curvature is the sum of all sectional curvatures of
planes spanned by pairs of orthonormal basis elements.
(1) The key is that $u$ is positive (as the exponential of something). As you already noted, $u$ is an eigenfunction of the Laplacian. But the strong maximum principle implies that (on a compact manifold) the only positive eigenfunctions of the Laplacian are the positive constants.
(2) The literature now contains far more sufficient conditions for the triviality of an $m$-quasi Einstein metric than I can keep track of. If you have access to MathSciNet, you can find all published papers which cite our paper, many of which contain some new sufficient condition. You can also use Google scholar for a similar purpose. As far as I know, only in dimensions one or two do we know exactly which Riemannian manifolds are $m$-quasi-Einstein; see this article of He, Petersen and Wylie.
Best Answer
In your setting, "constant curvature" means constant sectional curvature. The usual definition is "every tangent $2$-plane has the same curvature"; the formula you give for the Riemann curvature tensor is then a theorem.
In differential geometry, a metric with constant Ricci curvature is usually called an Einstein metric.
Being Einstein is not equivalent to having constant curvature. Up to overall scaling, there are precisely three local models for a space of constant curvature (and three complete, simply-connected examples: the sphere, Euclidean space, and hyperbolic space). By contrast, there are moduli spaces of Einstein metrics even on a single smooth manifold. For example, the smooth $4$-manifold underlying a complex quartic in complex projective $3$-space admits, up to scaling, a $20$-dimensional family of Ricci-flat Kähler metrics by Yau's solution of the Calabi conjecture.