I copy-paste here the exercise for completeness:
If at some point $P$, the symmetric tensor $R_{ab}$ satisfies
$$R_{ab}v^{a}v^{b} = 0$$
for an arbitrary timelike vector $v^{a}$, then deduce that $R_{ab}$
must vanish at $P$.
[Hint: let $v^a = t^a + \lambda s^a$, where $t^at_a = 1$,
$s^as_a = -1$, $t_a s^a = 0$, $0\leq\lambda<1$, $\lambda$ arbitrary, and
consider a special coordinate system in which $t^a=\delta^{a}_0$ and
$s^a=\delta^a_{1,2,3}$ in turn.]
So let's go carefully through the assignment. This exercise is a about a (generic) symmetric two-index tensor $R_{ab}$ (that he then applied to the Ricci), which vanishes contracted with timelike vectors (i.e. with vectors $v^a$ such that $v^av_a > 0 $); it is a general identity about tensors rather than a peculiarity of the Ricci.
Let's now follow the hint. As you correctly state, you get that $v_a v^a = 1- \lambda^2$ (this vector does not have unit norm, and there is no need why it should have); in particular it remains timelike for all values of $\lambda$.
Then we just compute what the contraction of $v$ with $R$ is:
$$ R_{ab} v^a v^b = R_{ab} (t^a + \lambda s^a ) ( t^b + \lambda s^b) = R_{ab} t^at^b + 2 \lambda R_{ab}t^a s^b + \lambda^2 R_{ab} s^a s^b,$$
where, in the last step, we used the symmetry of $R_{ab}$ and we renamed the two indices. Now we compute the previous expression for $t^a=\delta^{a}_0$ and $s^a = \delta^{a}_i$, being $i = 1,2,3 $; we get
$$ R_{ab} v^a v^b = R_{00} + 2 \lambda R_{0i} + \lambda^2 R_{ii} . $$
Since we observed that $v^a$ is timelike for any $\lambda$, we know by hypothesis that this quantity vanishes, i.e.
$$ R_{00} + 2 \lambda R_{0i} + \lambda^2 R_{ii} = 0.$$
Now we recall that $\lambda$ can assume any value between $0$ (comprised) and $1$; this is enough to say that all the components $R_{00}$, $R_{0i}$ (and by symmetry $R_{i0}$ as well) and $R_{ii}$ are equal to zero. Starting with $\lambda = 0$, we get $R_{00} = 0$, which holds for any choice of $\lambda$; then the equation (restoring $\lambda \neq 0$) reduces to
$$ 2 \lambda R_{0i} + \lambda^2 R_{ii} = 0,$$
which can clearly be satisfied for any $0<\lambda <1$ only if the two $R$'s vanish.
We have to show now that the mixed-spatial component $R_{ij}$ vanish; this is achieved by considering $s^a = (\delta^a_i + \delta^a_j)/\sqrt{2}$. Going through the same steps, and making use of the already-know-to-be-vanishing components we get $ 0 = R_{ab} v^a v^b = \lambda^2 R_{ij}/2.$
When a tensor vanishes in one frame of reference, it vanishes in all; this is a general property that worth being discussed more. The change of coordinates is reflected on tensors in a linear nonsingular transformation: we contract indices with the jacobian of the transformation; being the change of coordinates a diffeomorphism, the jacobian is injective, and therefore a tensor vanishes in one system of coordinates if and only if vanishes in all.
For completeness, I underline why the hypothesis of the symmetry of $R_{ab}$ is necessary. If it were not the case, we could not simplify the contraction as we did; we would just remain with
$$ R_{ab} v^a v^b = R_{ab} t^at^b + \lambda ( R_{ab} + R_{ba}) s^a t^b + \lambda^2 R_{ab} s^a s^b, $$ and here we have nontrivial cancellations between antisymmetric components of $R$: indeed, following the same steps, we would end up with just the symmetric part of $R_{0i}$ (i.e.\ $R_{(0i)} = (R_{0i}+R_{0i})/2$) vanishing. (But I am not 100% sure of this very last comment since it is quite late in the night and I had better to go to sleep)
