What does "surface area of a sphere" actually mean (in terms of elementary school mathematics)? I know what "surface area" means for:

*

*a 2d shape

*a cylinder or cone

but I don't know what it actually means for a sphere.
For a 2d shape
Suppose I'm given a 2d shape, such as a rectangle, or a triangle, or a drawing of a puddle. I can cut out a 1cm by 1cm piece of paper, and trace that piece of paper on the shape. Many full 1 cm squares will be traced on the shape, and there will likely be many partial squares traced on the edges of the shape. Suppose I can accept that I can "combine" the partial squares into full squares. Then I count the total number of full squares, to find the surface area.
For a cone or cylinder
I can convert a paper cone into two 2d shapes. The bottom of the cone is a circle. I can then cut the curved (ie not-bottom) part of the cone using scissors, and unfold that part into a flat 2d shape.
Similarly, I can convert a cylinder into flat 2d shapes: two circles and a rectangle.
For a sphere
But the above methods for understanding surface area don't work for a sphere. I can't lay a 1 cm by 1 cm piece of paper onto a sphere in a flat way. I can't even trace a square centimetre onto the sphere using that piece of paper!
People might say, "suppose you have an orange, and you peel the orange. Then you can lay the peel flat onto the table, into a flat 2d shape". But they're lying! The orange peel can never be mashed down perfectly flat onto the table!
So, I don't know what "surface area of a sphere" even means, if you cannot measure it using flat square pieces of paper!
What does "surface area of a sphere" even mean?
 A: There is a conceptually simple way to think of this: Build a hollow sphere out of some rigid material such as metal, or plastic. This material will have some thickness, say $d$. Suppose its inner radius is $r$ and its outer radius is $R$ (so we have $R=r+d$).
Now get out your kitchen scales and weigh the thing. Suppose it mass is $W$ grams; and suppose further that the weight of a unit square of your stiff material is $w$ grams. Then the surface are of the sphere is about $W/w$.
I say "about", because of the finite thickness $d$ of the spherical shell. But we know that the inner surface area is less than $W/w$ and the outer surface area is greater than $W/w$. And in the limit, as the thickness $d$ tends to zero, this value $W/w$ will tend to a limit, which is the area of the outer curved surface.
A: This is actually an interesting question. It involves how to define "area" on a curved surface. The examples you have provided are surfaces that are developable (can be flattened onto a plane) after a few cuts. And you can compute the flattened area. You can never do this to a sphere, because no matter how small a patch from a sphere is, it can never be flattened onto a plane. The idea is to break down the sphere to small patches such that each is flat enough and you compute the area as if it is flat, and then add up the areas of the patches.
Mathematically, suppose $S$ is a sphere. The above procedure is stated as:


*

*Break up $S$ into patches $P_1,\dots,P_n$, where each $P_i$ is a patch that is flat enough, and $n$ is the number of patches you have.

*Compute $\operatorname{Area}(P_i)$ as if each $P_i$ is flat. As suggest by levap, one way to do it is to project each patch onto one of its tangent planes. Note that I am not saying this is the only way to approximate a patch, and I am also not saying that one way that would seem correct at first glance would really be correct, see Update 2 for an example, there's also discussion about this in the comments.

*Use $\operatorname{Area}(P_1)+\dots+\operatorname{Area}(P_n)$ as an approximation of the area of $S$.

*If the patches are small enough, then the approximation should be a good one. But if you want better precision, use smaller patches and do the above again.

*This is to make the math precise, I can't guarantee that a third-grade student can understand this: As you take smaller and smaller patches, the value of the approximation above should tend to a fixed number, which is the mathematical definition of the area.
P.S. For a visualization of this approximation, you can search online for sphere parametrization, or simply think of a football (soccer ball).

Update 1: Thanks to Leander, we have a visualization:

One might notice that this visualization is slightly different from cutting up a sphere; it takes sample points on the sphere and attach triangles to these sample points. I want to remark that there is no essential difference between this and my method. The idea is the same: approximation.

Update 2: A comment (by Tanner Swett) mention that the method of using a polygon mesh may be flawed. Indeed, the example of Schwarz lantern shows that some pathological choice of the polygon mesh may produce a limit different from the surface area. The following explanation should be helpful:
As I have mentioned in step 2 above, if we are not careful with how we approximate the areas of the patches, the approximation may not work. The Schwarz lantern is an example where a careful choice of the approximating triangles can lead to the following result: Suppose $T$ is a triangle we use to approximate a patch $P$, then it is possible ${\rm Area}(T)/{\rm Area}(P)\to a\neq1$. To illustrate this, consider a single triangle on the Schwarz lantern:

We assume the cyclinder has total height $1$ and radius $1$. We take $n+1$ axial slices, and on each slice $m$ points. The area enclosed by the red curves is a patch on the cylinder, and the triangle enclosed by the blue dashed lines is the one used to approximate the patch. Let $P$ and $T$ denote the patch and the triangle respectively. We see that the bottom edge of $P$ and $T$ has ratio $1$ as $m\to\infty$. What really makes a difference is the ratio of their heights. Suppose along the vertical direction the height of $P$ is
$$h=1/n$$
Then the height of the triangle is
$$h_T=\sqrt{1/n^2+a^2}$$
By a simple computation we know $a=1-\cos(\pi/m)\approx(\pi^2/m^2)/2$. Therefore,
$$h_T/h=\sqrt{1+\frac{\pi^4n^2}{m^4}}$$
If $n$ has higher order than $m^2$, then the limit is bigger than $1$, and consequently ${\rm Area}(T)/{\rm Area}(P)\not\to1$.
This problem would have a smaller probability of occurring in practice. Imagine if you do cut the cyclinder into patches, you'd use $h$ instead of $h_T$ to estimate the area. But again, it is hard to make this (what approximation is acceptable) precise without using the language of calculus.
A: Consider a "convex polyhedron":

You can start with a simple pyramid or cube, but, as the polyhedron gets more and more complex, it can be made more and more like a sphere.  At each step along the way you can measure the dimensions of each flat surface, add the surface areas together, and come up with an estimate of the surface area of the equivalent sphere.  As the polyhedron is made with more and more pieces, it becomes a closer approximation to the sphere.
There is this mathematical concept known as a "limit" where the approximation, after an infinite number of refinements, essentially becomes a sphere, and the sphere's surface area is determined.
A: Take a sphere (or any other shape), and paint it blue. The amount of paint required is just proportional to the surface area. This is a way to measure it.
A: If the surface area of a sphere is $1\text{cm}^2$, that means that if you cut a sphere into very very very tiny pieces, so tiny they are almost perfectly flat, then the total area of those pieces will be very very very close to $1\text{cm}^2$.
A: You already got great answers. I wanted to emphasize that already for flat surfaces you are accepting to approximate your area by small rectangles. And I think it's clear to you that there will always be a small error, that you can diminish but never get rid of (unless you do calculus, and that's one of its magical traits).
With the sphere it's not different really. The leap you need to make is to accept that, instead of "missing area" just by the sides of your rectangles, now you will be "missing area" by not being able to set your paper rectangles flush against the surface. But it should be clear that, the smaller the rectangle, the better the approximation. 
One visualization that might help is to draw a circle with some plotting app (Desmos, to name one) and start zooming in. You will see that the more you zoom, the more the circle looks like a line. With the sphere, a 3d version of that phenomenon happens.
A: First, I'm a new contributor, so try not to jump on me!  :-)
Secondly, the o/p has asked how this problem might be explored in terms of elementary school math. I'm sure, at least, that we've all been there! Maybe we can approach this in terms of elementary grade math?
I was taken with the idea suggested of painting the surface area, and working out how much paint was required to paint the sphere's complete surface. 
If we knew how much paint we started with, and how much we had left afterwards, we could calculate the surface area of the sphere if we measured the thickness of the layer of paint now coating the sphere.
We might go mad and measure the diameter of the sphere before painting it, and after painting it, in order to use good old elementary grade subtraction to calculate the added diameter of the sphere with its fresh coat of paint. That would tell us the thickness of the coat of paint.
How about then looking at the problem from a new point of view? Still with our pot of paint, how about we actually dunk the sphere in it, in order to coat it with paint? And doesn't that suggest an additional test? How would it be if we measured the amount (volume) of paint displaced by the sphere?
Perhaps the paint pot might be completely full, so that immersing the sphere in the paint would cause paint to be displaced from the pot, and thus it could be measured as it flowed into a measuring vessel held beneath the pot, so that the volume of liquid displaced by the sphere would thus be measured. That would also give us a measurement of the volume of the sphere, which must be equivalent to the volume of liquid displaced.
Seems to me I learned about Archimedes in elementary school! Our grade school teacher's favorite joke was that 'Eureka' is Greek for 'this bath is too hot'!
Once we know the volume of the sphere, along with such certain (measured) properties as its radius and its circumference, we can make some calculations of its surface area. Perhaps if we made a series of such experiments, with spheres of differing surface area, we could eventually use simple multiplication or division to arrive at the well-known formula of 4 Pie R Squared.
And nothing has to be flattened onto a plane.  :-)
A: A circle of radius $r$ has area $\pi r^2$ and perimeter $2\pi r$. If we run a very thin line of pencil around the perimeter of thickness $\delta$, the graphite area will approximate $2\pi r\delta$.
A sphere of radius $r$ has volume $\frac43\pi r^3$ and surface area $4\pi r^2$. If we cover the surface with a very thin layer of spray-paint of thickness $\delta$, the volume of paint lost from the can will approximate $4\pi r^2\delta$.
Note that in both cases there are two formulae, one for how much space is inside the shape, and how much of a different kind of space, with one lower dimension, is on the edge of the shape. Basically, the edge size is how quickly the interior size grows as the shape widens.
(Edited to link to somewhat more detailed explanations.)
A: This is a very good question, with very good answers, so I'll just chime in with a comment.  Some years ago, a researcher came to me asking how to compute the surface area of a coffee bean.  I responded that it is very hard to get a really good answer.  Like others have said, you need to get a triangulation of the surface, and then add the areas of the triangles.  But if there are lots of small bumps on the coffee bean, it is hard to get a good approximation.
A similar question is "how long is the coast of England", which was originally asked by Benoît Mandelbrot.  The trouble is, as you focus closer and closer in to the surface, the answer gets longer and longer.
Now if he had asked for the volume of the coffee bean, that would have been easy.  Dunk it in liquid, and see how much spills over.  I could have told him to paint the coffee bean, and see how much paint he had to use, but it is hard to apply an even coat when the surface is bumpy.
A: Imagine a perfect sphere the size of the Earth, perfectly smooth, and that you've got a vast number of perfect little centimeter-square tiles and a large army of bored quarantined kids to lay them out and count them. 
On that huge sphere, each tiny tile will seem to lay flat, and to fit perfectly with the tiles on all four sides, and cover the planet with no visible gaps; and after you tally them all you can say that the surface area of the earth is so-many square centimeters. It will be a very (very!) large number, but it'll be a definite number and that's the surface area. 
For a smaller sphere, like a beach ball or an orange or a ping-pong ball, a square-cm tile isn't going to fit well at all. So use a smaller tile: one a mm square, or a micron, or Angstrom, or smaller. Give your kids tweezers and magnifying glasses and get them to work. Eventually you'll have the surface area of your sphere, in sq mm, or square Angstroms, or barns (yes, that's a unit of area!) or whatever.
So to conceptualize the surface area of a curved surface, just think smaller and smaller until your hypothetical tile is so much smaller than the curvature of the surface that it seems to lie flat and join perfectly with the tiles surrounding it. And get ready to count to very large numbers.
A: I would first introduce approximating the area of a shape and pi via the method of exhaustion

The area or circumference is approximately the average of the two, but not quite..
Once students understand this for a two-dimensional shape, it should seem clear both


*

*pi exists and is a transcendental number

*it is illogical to try and represent the area or circumference of a circle without it


With this out of the way, you could pose using Exhaustion with an N-faceted polygon (perhaps beginning with a cube inside a cube?). Ideally this will lead them to discover again that they will need pi to find the true surface area, while also subtly preparing them for calculus.
Plausibly you could purchase or fashion an object to show this, but I suspect some graphics simulation software will aid you (and also trivialize discovering the area of the contained and surrounding solids)
A: All the solid shapes whose surfaces you are able to understand are finitely rectifiable -- that is, you can think of a finite number of transformations such that the areas (if we agree that areas are additive) can be transformed from covering a solid to lying entirely in one plane.
However, there's no reason to think that this will always be possible for all surfaces which clearly posses an area (albeit intuitively). That would be tantamount to the naive assumption of the Pythagoreans that all quantities can be measured using only integers and ratios of integers -- hence their intense shock on discovering the irrationality of the diagonal of a square!
The general lesson is that elementary methods aren't always sufficient to capture everything we would like to capture -- we have to extend our elementary methods and notions in a way that accommodates objects that wouldn't fit in the earlier scheme, while still preserving their logical character. This is precisely the triumph of the infinitesimal calculus over finite calculus. Many things may be done (with more and more difficulty) with only the latter, but sooner or later one has to admit that one cannot escape using infinitesimal analysis, for even some very basic things.
So, again, the point boils down to an extension, a rising I'll say, from finite methods to infinite methods. That one has to do this doesn't mean that the objects that only admit of infinite methods don't have the properties analogous to those objects that can be conquered using only finite methods -- after all, those old objects can be consistently analysed using the new infinite methods as well.
So, how to understand the surface  of a sphere? Accept that it may be impossible to rectify in only a finite number of steps, and accept therefore that you would need infinitely many operations to completely rectify it. Accept that this is not strange, since in the end you'll have a definite quantity for your area. Finally, since we have only finite brains, how do you think of this process -- just cut the sphere into smaller and smaller pieces (one way is to go along the longitudes), and continue ad infinitum. As you continue this process, you see that the strips become thinner and thinner, and so more and more rectifiable, although they still contain a tinge of curvature. This curvature will never disappear after any finite number of thinning steps, but it gets arbitrarily smaller, so that we know that it approaches a definite rectified form. This is the approach of limits. In the approach of infinitesimals, one would say that after infinitely many such operations the strips become infinitely thin, and flexible, so that the curvature may be completely removed.
Then the sum of the areas of all these strips, gives the area. In the limit approach, you'd have to approximate the area of each strip at each stage of the process, and note that the approximations get arbitrarily close to a certain quantity, which is the desired area.
A: Archimedes showed that the surface area of a cylinder (without the top and bottom) is equal to that of the inscribed sphere.  Further, the areas cut off by any planes perpendicular to the cylinder's axis are also equal.  This makes intuitive sense as follows.  The angle at which the sphere "recedes" at any "latitude" gives you MORE surface area than the cylinder slice.  However, the smaller radius of the slice of the sphere at that "latitude", gives you LESS surface area than the cylinder slice.  By drawing some triangles, I was able to convince myself that the MORE and the LESS offset each other exactly.  

