Is $ds^2$ just a number or is it actually a quantity squared? I originally thought $ds^2$ was the square of some number we call the spacetime interval. I thought this because Taylor and Wheeler treat it like the square of a quantity in their book Spacetime Physics. But I have also heard $ds^2$ its just a notational device of some sort and doesn't actually represent the square of anything. It is just a number and that the square sign is simply conventional. 
Which is true? 
 A: It is a mnemonic notation that indicates that $\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$ is the object whose square root is to be used as the infinitesimal line element, traditonally denoted $\mathrm{d}s$, when determining the lengths of worldlines $x : [a,b] \to \mathcal{M}$ by integrating the line element along them as
$$ \begin{align*}
L[\gamma] & = \int^b_a \lvert\lvert x'(t) \rvert\rvert \mathrm{d}t = \int_a^b  \sqrt{\lvert g_{\mu\nu} x'^\mu(t) x'^\nu(t)\rvert}\mathrm{d}t = \int_a^b \sqrt{\lvert g_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}t}\frac{\mathrm{d}x^\nu}{\mathrm{d}t}\rvert}\mathrm{d}t \\ & = \int_x\sqrt{\lvert g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu\rvert}\frac{\mathrm{d}t}{\mathrm{d}t} = \int_x\sqrt{\lvert g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu\rvert} = \int_x \mathrm{d}s 
\end{align*} $$
where the left hand side defines the length functional and the right hand side is obtained by scetchy manipulation of differentials, which is why you should not take $\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$ too seriously.
A: As already mentioned by others, $\mathrm ds^2$ is used as suggestive notation for the metric tensor
$$
g = \sum_{\mu,\nu}\mathrm g_{\mu\nu} \, dx^\mu\otimes\mathrm dx^\nu
$$
In case of a positive definite metric and given a curve $\gamma:[0,T]\to M$, it has a precise meaning in terms of either the length function
$$
s_\gamma(t) = \int_0^t \sqrt{g(\dot\gamma(\lambda),\dot\gamma(\lambda))}\;\mathrm d\lambda
$$
with derivative
$$
\mathrm ds_\gamma = \sqrt{g(\dot\gamma,\dot\gamma)}
$$
or equivalently in terms of the pullback
$$
\gamma^*g = \mathrm ds\otimes\mathrm ds
$$
where $s$ denotes the induced normal coordinate on the interval.
A: It is a notational device. Note that in $(-+++\cdots)$ the proper length
$$ds^2=g_{\mu
\nu}dx^\mu dx^\nu$$
is negative for timelike $dx$. Thus $ds\equiv \sqrt{ds^2}\in\mathbb{C}$. It (the square root) thus has no physical meaning. 
A: It is a square of a proper time interval or a square of proper distance (modulo an inessential sign).
A: If you use the $({-}{+}{+}{+})$ sign convention then $ds$ is the proper distance between two infinitesimally separated points; if you use the $({+}{-}{-}{-})$ convention then it's the proper time. In each case, if you choose points whose separation is such that the proper distance (respectively, proper time) is not meaningful then $ds$ will become imaginary.
On purely aesthetic grounds, I've always felt that proper time is more fundamental than proper distance, since it has a direct physical effect and can measured, while proper distance must be inferred. From that perspective I find it nicer to use the $({+}{-}{-}{-})$ convention and think of $ds$ as an interval of proper time, which becomes imaginary if no particle can travel along the path in question.
But from the point of view of actually doing calculations, none of this matters. $ds$ always appears in its squared form, so whichever sign convention you choose, if you're bothered by the idea of an imaginary time or an imaginary distance you can avoid it by treating $ds^2$ rather than $ds$ as the fundamental quantity. Neither the choice of sign convention nor the choice of how to interpret $ds^2$ changes the result of any calculation, so they make identical physical predictions and are ultimately both matters of taste.
