Since the contraction of two vectors need them both to be defined at the same point $p$ and we are free to choose whichever coordinates we prefer on the manifold, we can always choose normal coordinates at $p$, so that the problem ends up reducing to the same thing in Minkowski spacetime.
In a more physics-like language, we can do the computation in a free-falling frame so that the computation is done as in Minkowski spacetime.
Let us choose the null vector $n^a = \left(\frac{\partial}{\partial t}\right)^a + \left(\frac{\partial}{\partial x}\right)^a$ (you can check it is indeed lightlike). The vectors $t^a_\pm = \pm \left(\frac{\partial}{\partial t}\right)^a$ are timelike. Notice now that, under the $-+++$ convention,
\begin{align}
n_a t^a_\pm &= \pm \left(\frac{\partial}{\partial t}\right)^a \left(\frac{\partial}{\partial t}\right)_a + \pm \left(\frac{\partial}{\partial t}\right)^a \left(\frac{\partial}{\partial x}\right)_a, \\
&= \mp 1 + 0, \\
&= \mp 1.
\end{align}
Hence, the contraction of a timelike vector with a null vector can be either positive or negative. A similar result works for spacelike vectors, and can be proven in a similar way.
In the $-+++$ convention, it does hold that the contraction of a timelike and a null vector will be negative if they both have the same time-orientation (i.e., if both point to the future or both point to the past). To see this, pick coordinates such that the timelike vector is of the form $\left(\frac{\partial}{\partial t}\right)^a$ (I'll assume the timelike vector to be normalized for simplicity). Completing the coordinate system so that it is normal, we can write the most general vector as
$$n^a = \alpha \left(\frac{\partial}{\partial t}\right)^a + \beta \left(\frac{\partial}{\partial x}\right)^a + \gamma \left(\frac{\partial}{\partial y}\right)^a + \delta \left(\frac{\partial}{\partial z}\right)^a.$$
Imposing $n^a n_a = 0$ now leads to
$$-\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 0,$$
and hence $n^a$ is given by
$$n^a = \pm \sqrt{\beta^2 + \gamma^2 + \delta^2} \left(\frac{\partial}{\partial t}\right)^a + \beta \left(\frac{\partial}{\partial x}\right)^a + \gamma \left(\frac{\partial}{\partial y}\right)^a + \delta \left(\frac{\partial}{\partial z}\right)^a.$$
If $n^a$ and $\left(\frac{\partial}{\partial t}\right)^a$ both point to the same time direction, then the $+$ sign is to be chosen. Otherwise, the $-$ sign. Assuming $+$, we see that
$$n_a \left(\frac{\partial}{\partial t}\right)^a = - \sqrt{\beta^2 + \gamma^2 + \delta^2} < 0,$$
where the strict negativity comes from the fact that if $\sqrt{\beta^2 + \gamma^2 + \delta^2} = 0$, then $n^a$ would be timelike instead of null.
