This is a great question, with three great answers, and here I am, a bit late to the party. There are two crucial things which the above answers don't seem to address, so I am going to try to give a really simple explanation at those levels.
Locality / Manifolds
I'm going to come close to giving you one technical definition of a manifold, following Penrose's book Spinors and Space-Time, but I will try to keep this light and common-sense. Let's start by forgetting what else we know about space for a second and focus on one thing: spacetime is a set of "points". Sure, it's not a "discrete" set (it has an uncountably infinite number of points in it), but spacetime is fundamentally defined by the fact that it contains these other objects, which are "points" in spacetime, and we can't really define those points further. They're just some sort of atomic thing that we can talk about, and spacetime is a set $\mathcal S$ of those points.
Obviously, without any more information we're pretty much stuck.
Arrows as information
The simplest form of information is to create the allowable scalar fields $\mathcal A \subseteq (\mathcal S \rightarrow \mathbb R)$, which means "$\mathcal A$ is some subset of (or equal to) the functions which take points in $\mathcal S$ to numbers in $\mathbb R$." We could call these "scalar fields" but maybe an even better thought is to steal some category theory and call them "arrows". By restricting the arrows we choose in the set $\mathcal A$ we can actually tune how fussy/persnickety we're being about our descriptions of $\mathcal S$, since they're our only tool to describe $\mathcal S$. It is nonetheless good to make some really basic assumptions. One broad assumption is about closure: suppose $f$ is a smooth function $\mathbb R^n \rightarrow \mathbb R$, then it makes sense that if $a_1, \dots a_n$ are all arrows (scalar fields in $\mathcal A$), then the scalar field $p \rightarrow f(a_1(p), \dots, a_n(p))$ should also be in $\mathcal A$. I'm going to use these smooth-functions again, so let's write $p \rightarrow f(a_1(p), \dots, a_n(p))$ as $f[a_1, \dots, a_n]$ with square-brackets, to remind ourselves that we're "lifting" the definition of $f$ from $\mathbb R^n \rightarrow \mathbb R$ to $\mathcal A^n \rightarrow \mathcal A$ in the "most obvious" way we can.
That closure axiom gives us useful operations like $+$ and $\cdot$ on arrows, "lifting" those operations from reals to scalar fields in $\mathcal A$. In fact the closure of $\mathcal A$ gives us something even deeper: the "kernel closed-set topology". Define a set $s \in \mathcal S$ as "closed" if there exists an arrow $a \in \mathcal A$ such that $\operatorname{ker} a = s$ (in other words, $s$ is the set of points that $a$ takes to the number $0$). Define a set as open if its set-complement is closed. This set of definitions is called a topology (if it meets certain axioms, which the closure axiom guarantees) and it lets you define what it means for maps to be "continuous" (which, it turns out, all arrows are). It does this with some other definitions: for example a "neighborhood" of a point is an open set which contains that point.
So if $\mathcal S$ is a 2-sphere (i.e. a surface of a 3-ball), the obvious thing is to use 3D coordinates and assign values $x$, $y$, and $z$ to the sphere, allowing any smooth functions $\mathbb R^3 \rightarrow \mathbb R$ of those three numbers to be our "arrows".
Coordinates as arrows
And now we can finally talk about what it means to have coordinates by adding another axiom on $\mathcal A$: for every point $s \in S$ then it has a neighborhood $N_s$ and some arrows $c^s_1, ... c^s_d$ such that if two points $p, p' \in N_s$ are different ($p \ne p'$), then some coordinate is different ($c^s_i(p) \ne c^s_i(p')$ for some $i$). So the arrows can be used, locally, to tell points apart.
Notice that we can do this with 2 coordinates on the 2-sphere -- we can do it locally, but not globally. ($\theta$ is not an arrow; it has a discontinuity.) So if a point is in the Northern or Southern hemisphere we can use $x, y$ as coordinates. If it's on the equator we either need to use $x, z$ or $y, z$ as coordinates. There are continuous functions which map $x \le 0$ to $0$ and so forth, so that's all we need: the neighborhoods are hemispheres.
This gives a very technical but 100% valid reason for "why do we need coordinate-free descriptions": once you know what coordinates are, the fact that they are local to certain points in spacetime means that any coordinate-specific definition is only valid in a neighborhood of your current position. You start giving a 2D description of your world in terms of North/South and East/West, but your description fails when you start going more than 10,000 km East or West. Once you define your function well enough, it becomes an arrow and therefore "coordinate-free".
Similarly you often see people say that "it'd take forever to reach a black hole" or "time stands still at the event horizon" or similar things. We've actually known for a long time now that that's not true, and it's an error of the same sort. If you're standing still and you're far-away from the black hole, you have certain coordinates which are "natural" for describing it. It turns out that your coordinates can't pass through the event horizon of the black hole. But that doesn't mean that matter can't. In fact, we discovered that there are "co-moving coordinates", a set of coordinates used by someone who is falling naturally into the black hole, which work just fine for getting across the event horizon of a black hole. It's just that when you fall past the event horizon at your local time $t$, you stop being able to send light of yourself to the person who is far away from the black hole: so the last image that they see of you is at your time $t$, and because the light from that time needs to escape more-and-more gravity the closer you get to the horizon, images from that time take longer and longer to get there. From their perspective, then, of course it sort of "looks like" it takes you forever to reach the black hole and like everything at the event horizon "stops changing" and all that. But that's because their coordinates can't pass through it.
Vector fields make it even more obvious
A different example: you can certainly start to define vector fields now on our $(\mathcal S, \mathcal A)$ topological-space, as derivations: functions from arrows to arrows which satisfy a "chain rule" in the sense that, if $f$ is a smooth function $\mathbb R^n \rightarrow \mathbb R$ (remember those?) where $f_(i)$ is the first derivative of $f$ with respect to its $i^{\text{th}}$ parameter, then $$\mathbf v(f[a_1,\dots,a_n]) = \sum_{i=1}^n f_{(i)}[a_1, \dots, a_n] \cdot \mathbf v(a_i).$$ Notice again that this means that $\mathbf v(a + b) = \mathbf v(a) + \mathbf v(b)$ and $\mathbf v(a \cdot b)$ = $a \cdot \mathbf v(b) + \mathbf v(a) \cdot b$. We get a lot from these smooth functions $\mathbb R^n \rightarrow \mathbb R$! (If you've never seen a vector field as a derivation before, in Euclidean space with your usual orthonormal coordinates, inner product $\langle\cdot,\cdot\rangle $ and vector fields, the derivation corresponding to field $\vec v$ is the function $\mathbf v(f) = \langle \vec v, \nabla f \rangle$.
Now, locally, you can define nonzero-everywhere vector fields on your coordinates, fitting those rules: but on the 2-sphere, there are no vector fields which are nonzero everywhere. This has the charming name of the "hairy ball theorem" and if you interpret the vector fields on the 2-sphere as wind profiles, it says that there are essentially always at least two "cyclones" (measured by counting eyes-of-storms where the wind speed is 0) on the sphere.
You wouldn't have expected this with your coordinate-based descriptions of the wind, would you?
Generalized coordinates
It's actually not just manifolds and general relativity where you benefit from coordinate independence: much earlier in their school time we start teaching students about variational calculus and generalized coordinates. The usual way to motivate these is the Brachistochrone problem:
Consider a purely-gravity-fed friction-free roller-coaster starting at $(-L, 0)$ and ending at $(L, 0)$. Find the track $y(x)$ that the roller coaster has to take, so that the trip from start to finish takes the least amount of time.
We can see two extremes. Think of a track that goes straight down some very height $h$, followed by a sharp turn that converts it all into forward momentum at $(-L, -h)$, followed by a sharp turn that converts it all into upward momentum at $(L, -h)$. From the energy, we know that the speed at the bottom of this track is $v^2 / 2 = g h$ or $v = \sqrt{2 g h}$. The free-falls speed linearly up from $0$ to $h$, so they average half this speed, $\sqrt{g h / 2}$. So the total time for this sort of track is$$T = \frac{2 h}{\sqrt{g h / 2}} + \frac{2 L} {\sqrt{2 g h}} = \frac {4 h + 2 L} {\sqrt{2 g h}}.$$
The two extremes here are $h \rightarrow 0$ where $T \rightarrow \infty$ (no free-fall, but also not enough forward speed) and $h \rightarrow \infty$ where $T \rightarrow \infty$ also (moving infinitely fast, you cross the $2L$ distance in no time -- but you take forever to fall far enough to get there. With some calculus you can even find a minimum $T$ between those, assuming that form of the path -- but there are other, more-curvy paths to investigate between these straight lines.
The problem is that those "sharp turns" in the above example are the only constraint forces which are "nice". A constraint force has to operate in tandem with a particle's momentum, pointing that momentum into the allowed-directions. And if you don't yet know the track then you don't yet know the direction of the constraint force -- much less the magnitude! So the most important force to this problem (in fact, the only force other than gravity) is actually pretty hard to figure out even if we assume that you have a form for $y(x)$. It's hard to do this problem with classical, coordinate-based methods and forces and such.
Something similar happens for double-pendulums: the constraint force that keeps the two parts of the pendulum at the same distance is varying its direction constantly; how are you going to handle it? Well, wouldn't it be nice if you had a coordinate-independent understanding of the physics so that you could choose coordinates which enforce the constraints -- in this case $\theta_1, \theta_2$, and then do physics with those? After all, if you choose the right coordinates then you don't have to model the constraint force. And then you can get some differential equations for the double pendulum, prove that it's chaotic, and build two of them to show chaos to your classroom.
