How does the surface area of a rigid square piece of a metal traced on a sphere compare with the area of that piece of metal? My main question is in Scenario 2. I don't know how to answer this main question.
Scenario 1 asks a similar question that I am able to answer, that led me to think of my main question.
Scenarios 3 and 4 ask similar questions to my main question, that I attempted to try to help me answer my main question.

Scenario One -- laying down a flexible piece of paper onto an sphere, and tracing the resulting shape
Suppose I have a 1cm x 1cm piece of paper, and a sphere that is somewhat larger (for example, a sphere about the size of an orange).
Suppose I lay the square piece of paper down on the orange, and I flatten it as much as I can. It is impossible to do this in a way where all of the piece of paper will touch the orange; some parts might be crumpled "into the air" where air is in between the paper and the orange; and some parts might be lying close to the orange, but paper is in between that part and the orange.
Suppose I then use a maker to trace a shape onto the orange, by tracing around where the edges the piece of paper touch the orange.


*

*A question: how does the surface area of this traced-out shape compare with the area of the square piece of paper?

*My answer: Intuitively, I can visualize that the surface area of the shape traced onto the orange will be less than the area of the square piece of paper, because only some points of paper are directly touching the orange; there are other "extra" points of paper that are in the air or lying on top of other points of paper. These "extra" points of paper are not contributing to the shape that is traced out, therefore the shape that is traced out must be less than the area of the square piece of paper.


Scenario Two -- rolling a rigid piece of metal onto a sphere, and tracing the resulting shape
Suppose, now, that I have a rigid square piece of material (such as a piece of metal) that cannot be bent, that is 1cm x 1cm, and a sphere that is about the size of an orange.
Suppose I use this piece of metal to trace a shape onto the orange in the following way:


*

*I start by placing the centre point of the square onto the orange, such that it's the only point touching the orange.

*Then I smoothly roll the square piece of metal along the orange, making a line from the centre of the square piece of metal, to the midpoint of one the edges of the piece of metal. I then mark a dot onto the orange, with a marker, where the midpoint of the edge of the square touches the orange. I then undo this rolling motion, so that the centre point square piece of metal is once again touching the orange.

*Similarly, I mark the midpoints of the other three edges of the square (ie, starting from the centre, and rolling it to those midpoints). And similarly, I mark the corners of the square. And then, I mark many, many other points along the edge of the square.

*Finally, I can join all the points marked with some sort of smooth (possibly curved?) line.


Main Question: A shape now is "traced" onto the orange. Is its surface area less than, equal, or greater than 1 square centimetre?
I have trouble even making a guess! (I welcome any observations that may help me visualize certain properties of this scenario, even if they don't provide a mathematically rigorous explanation. I especially welcome explanations that use high-school (or earlier) level mathematics.)
Indeed, I can't even make a guess about what shape is traced! Are the corners of the traced shaped right angles? If looking directly down at the shape (ie from a bird's-eye-view, as if your eye, the centre of the shape, and the centre of the sphere, made a straight line), would the shape look like a square? I can't visualize this!

Similar Questions I asked myself, as attempts to gain an understanding of the Main Question
Scenario Three -- tracing a rigid square onto a cylinder
Suppose a trace a shape onto a cylinder in the following (obvious) way:


*

*I lay a cylinder flat on a table (ie, so one of it's circular faces is resting on the table).

*I then place one edge of the square so that it is perpendicular to the table, against the side of the cylinder, and I draw a line along that edge, with marker, onto the cylinder.

*Then I smoothly roll the square towards the other vertical edge of the square. As I smoothly roll the square, the horizontal edges of the square will "roll along" the cylinder. As they do so, I mark the places where they contact the cylinder.

*Finally, I draw a vertical line onto the cylinder, when I reach the square's other vertical edge.


I am somewhat confident that the surface area of the resulting shape would equal one square centimetre:


*

*It seems that (as I'm rolling the square along the cylinder) if I take any two points on the square that are vertical to each other, the distance between these two points on the square will be equal to the distance between the points where they each touch the cylinder (as the square is being rolled onto the cylinder).

*Also, suppose I consider any two points that are horizontal to each other on the square, and I mark where these points touch the cylinder (as the square is being rolled), using purple ink on the cylinder. If I measure the distance between the two points on the square, and if I measure the distance an ant would walk along the curved surface of the cylinder between the two points in purple ink that I marked on the cylinder, my guess is that these two distances would be the same.


Considering this, it feels like all points of the square will be "placed" onto the cylinder's surface without any distortion/warping/stretching. That's why I am making a guess that the surface area of the resulting shape traced onto the cylinder will equal the surface area of the square piece of metal.
Scenario 4 -- Tracing a square onto a cone
I am not confident in my guess here. My guess is that the shape traced onto the curved part of the cone will have a surface area equal to the area of the square.
However, I have trouble even visualizing the tracing process.
Suppose I start with the cone resting on the table (ie, with the circular face of the cone flat on the table). I note that I can choose a line on the curved surface of the cone, that runs from the tip of the cone down to any point on the edge of its circular face. I then start by placing the square onto the cone, such that the midpoint of two opposite sides of the square align with the "line" on the cone that I chose earlier. I can then do an obvious "rolling" motion of the square, once going right, and once going left, to trace out a shape onto the cylinder.
But I am unsure if I can visualize the shape created:


*

*For example, when rolling the square, starting with the starting placement described above, and then rolling right, there will be a line traced out (onto the cone) by the upper horizontal edge of the square, and a line traced out by the lower horizontal edge of the square. Will these lines that are traced out be parallel to the table? My intuition is saying "no"; my guess is that that the lines will slant downwards towards the table, as you roll the square from the middle of the square to its right edge .. but I'm not sure.

*Similarly, the vertical(ish) line created by the vertical right edge of the square .. will it be "vertical" (ie, aligned with a line that runs from the tip of the cone, down to a point the edge of the cone's circular face)? I cannot visualize this well enough to even make a guess.

 A: I think the area would shrink in the same manner the piece of paper had. If you think about it, in the paper case, the distances of the points of the perimeter of the square to its center are not different between the flat and the curved square. Once projected on the sphere, every line you trace on the paper that connects the center to its perimeter, still has the same length if you remove it from the sphere and flatten it again. So the paper one is a process that does not change the length of radial lines but does it for every other line. The thing would be different if the paper process or the metal one changed radial distances, but I'm assuming it doesn't and with that assumption the two projections would be equivalent imo. Hope I've been clear and tell me what you think about it, cheers :)
A: You have done an excellent job of describing your problem!
Scenario One is clear: the resulting area is less than 1 cm$^2$, because of the wastage in the paper.
Scenario Two is identical to Scenario One, if I have understood your procedure correctly. The boundary of the shape will be the same in both cases.
Scenarios Three and Four are different. Both a cylinder and a cone can be cut along a straight line and flattened into a plane. So in both these cases, the resulting shape will be unchanged from the original 1cm square.
One way to categorise the difference is to say that your orange has positive intrinsic curvature, while the cylinder and the cone have zero intrinsic curvature (i.e. they are intrinsically flat). Some surfaces (a torus, for instance) have negative intrinsic curvature; and on these, the area will actually increase.
A: The traced area of metal square of side $1$ cm will be exactly $1\text{ cm}^2$ on both the cylinder and cone because as indicated by @TonyK (and also by @Narasimham) they can be flattened.  So in this answer I will explain what I did to find out what the traced area of the square on a sphere will be.
Imagine a sphere of radius $R$ and place a square of side {2 a} on top of it (the north pole).  Now tilt the square (as described by @silph) in a direction making a angle $\phi$ with the $x$ axis, where $0 \lt \phi \lt \dfrac{\pi}{4} $, then the radial distance from the center of the square to its perimeter along this direction is
$ r = a \sec \phi $
Hence, the angle made with the $z$ axis is $\theta_0 = \dfrac{a}{R} \sec \phi$
That's all the necessary math, at the final step, I set up a double integral in $\phi$ and $\theta$ to find the area, taking symmetry into account introduces a factor of $8$.
$\text{Area} = \large 8 \displaystyle \int_{\phi = 0 }^{\phi = \frac{\pi}{4} } \int_{\theta = 0 }^{\theta = \theta_0 } \sin (\theta) R^2 d\theta d\phi$
Using $a = 0.5$  (half the side length) , and $R = 5$ (sphere's radius) results in
$\text{Area} = 0.998889$
Using $a = 0.5$ , and $R = 20$ (bigger radius) results in
$\text{Area} = 0.999931$
And finally using $a = 0.5$ and $R = 2$ (small radius) results in
$\text{Area} = 0.993076$
A: Here's a description of the figure obtained by moving the metal square over a sphere of radius $\ r_s\ \big(\,>\frac{1}{\pi\sqrt{2}}\big)\ $ (your  scenario $2$), and a proof that its area is less than that of the square.  The figure will comprise $8$ congruent octants, each generated from one of the octants of the square.  It is sufficient to describe one of these octants and show that its area is less than $\ \frac{1}{8}\,\text{cm} ^2$.
I take the sphere to be centred at the origin of a spherical polar coordinate system, with the starting position of the square having its centre at the point $\ r=r_s,$$\ \theta=$$\phi=0\ $, and its corners at the points $\ r=\sqrt{r_s^2+\frac{1}{4}}$, $\phi=\arccos\frac{1}{2r_s},$ $\ \theta=\frac{\pi}{4},$ $\frac{3\pi}{4},$ $\frac{5\pi}{4},$ $\frac{7\pi}{4}\ $.  That is, in the corresponding Cartesian coordinate system, the square lies in the plane $\ z=r_s\ $ with its sides parallel to the $\ x\ $ and $\ y\ $ axes.
When you roll the square along the circular arc in which the plane $\ \theta=\nu\ $ (with $\ 0\le\nu\le\frac{\pi}{4}\ $) intersects the sphere, the length of the arc will be the same as the length of the line from the centre of the square to its edge in which the plane $\ \theta=\nu\ $ intersects it—namely $\ \frac{1}{2\cos\nu}\,\text{cm}\ $. This will subtend an angle of $\ \frac{1}{2r_s\cos\nu}\ $ at the centre of the sphere. Thus, the set of spherical polar coordinates of the octant generated on the sphere by that on the square lying between the half-planes $\ \theta=0\ $ and $\ \theta=\frac{\pi}{4}\ $ is
$$
A=\Big\{\big(r_s,\theta,\phi\big)\,\big|\,0\le\theta\le\frac{\pi}{4},\,0\le\phi\le\frac{1}{2r_s\cos\theta}\Big\}\ ,
$$
and the area of the octant is
\begin{align}
\iint_A r_s^2\sin\phi\ d\phi d\theta&=r_s^2\int_0^\frac{\pi}{4}\int_0^\frac{1}{2r_s\cos\theta}\sin\phi\,d\phi\,d\theta\\
&=r_s^2\int_0^\frac{\pi}{4}\Big(1-\cos\Big(\frac{1}{2r_s\cos\theta}\Big)\Big)\,d\theta\ .
\end{align}
I doubt if there's any simple expression for this  integral.  However, $\ \cos x>1-\frac{x^2}{2}\ $ except for $\ x=0\ $. Therefore, for all $\ \theta\in\big[0,\frac{\pi}{4}\big]\ $, we have
\begin{align}
1-\cos\Big(\frac{1}{2r_s\cos\theta}\Big)&<\frac{1}{8r_s^2\cos^2\theta}\ .
\end{align}
Therefore, the area of the octant
\begin{align}
\iint_A r_s^2\sin\phi\ d\phi d\theta&<\frac{1}{8}\int_0^\frac{\pi}{4}\sec^2\theta\,d\theta\\
&=\frac{\tan\big(\frac{\pi}{4}\big)-\tan0}{8}\\
&=\frac{1}{8}\ .
\end{align}
It follows that the area of the sphere rolled over by the square in scenario $2$ is strictly less than $1\,\text{cm}^2$.
Bird's-eye view
You can see what a "bird's-eye" view of the figure would look like by projecting it down on to the plane $\ \phi=\frac{\pi}{2}\ $—i.e. the $x$-$y$ plane, in cartesian coordinates.  If the distance of the view-point relative to the size of the
sphere is sufficiently large, you can assume that the projection is effectively vertical.  Now the image of a point with spherical polar coordinates $\ (r,\theta,\phi)\ $ projected vertically downwards is the point with polar coordinates $\ (r\sin\phi,\theta)\ $.  Therefore, a point on the outer boundary of the octant described above, with spherical polar coordinates $\ \Big(r_s,\theta,\frac{1}{2r_s\cos\theta}\Big)\ $, will project to the point with polar coordinates $\ \Big(r_s\sin\Big(\frac{1}{2r_s\cos\theta}\Big),\theta\Big)\ $.  The  projection of the octant's outer boundary is therefore described by the polar equation
$$
r=r_s\sin\Big(\frac{1}{2r_s\cos\theta}\Big)\hspace{2em}\Big(0\le\theta\le\frac{\pi}{4}\Big)\ .
$$
Rotating this curve clockwise through an angle of $\ \frac{\pi}{4}\ $ produces a curve with the same polar equation, but with $\ \theta\ $ in the range $\ \big[{-}\frac{\pi}{4},0\big]\ $. Adding the  curves obtained by rotating these two anticlockwise through angles $\ \frac{\pi}{2},$$\ \frac{3\pi}{2},$ and $\ \frac{5\pi}{2}\ $ produces the curve with polar equation
$$
r=\cases{r_s\sin\Big(\frac{1}{2r_s\cos\theta}\Big)& for $\ {-}\frac{\pi}{4}\le\theta\le\frac{\pi}{4} $\\
r_s\sin\Big(\frac{1}{2r_s\cos\big(\theta-\frac{\pi}{2}\big)}\Big)& for $\ \frac{\pi}{4}\le\theta\le\frac{3\pi}{4} $\\
r_s\sin\Big(\frac{1}{2r_s\cos(\theta-\pi)}\Big)& for $\ \frac{3\pi}{4}\le\theta\le\frac{5\pi}{4} $\\
r_s\sin\Big(\frac{1}{2r_s\cos\big(\theta-\frac{3\pi}{2}\big)}\Big)& for $\ \frac{5\pi}{4}\le\theta\le\frac{7\pi}{4}\ $.
}
$$
The figure represented by this equation is a quadrilateral whose sides bulge outwards.  Except when $\ r_s=\frac{\sqrt{2}}{\pi}\ $,  the corners where the sides meet are sharp.  Geogebra plots of the figure superimposed on a circle of radius $\ r_s\ $ are given in the diagram below for three different values of $\ r_s\ $.  The plots are not all drawn to the same scale.  When $\ r_s\ $ is fairly large, as for the case $\ r_s=4\ $ in the diagram below, the bulge in the sides of the figure is quite small, and the figure is effectively indistinguishable from a square. As $\ r_s\ $ decreases, the bulge in the sides becomes more pronounced, and the angle at the corners increases, making them less sharp.
When $\ r_s=\frac{\sqrt{2}}{\pi}\ $, the diagonal of the square is large enough to reach halfway round the circumference of the sphere. The figure no longer has any sharp corners in this case, because the angles at the points of maximum curvature have become perfectly straight.

A: The applicability of surfaces is studied after high school, at the university/college..
Draping a small square on a cylinder or cone is possible because the same zero Gauss curvature matches.
Full draping or application between a plane and spherical orange surface is not possible by Egregium theorem in Differential geometry. Attempting such a drape/cover involves less than 100% contact of given square area.
Draping a one square centimeter on the surface area of 100 mm sphere diameter is uninteresting to me due to small patch size.
However the following addresses an approximate procedure when $ a=100 mm X 100 mm $ like sizes of paper and is indicated to start a covering. It may not be fully satisfactory. Happy to answer fwiw due to your motivation and description of the task. Never mind the bounty.
Convert the given square to a circle of same area $ \pi r^2= a^2;$
Mark 25 mm circle dia at north pole.
Width at north pole $ \pi 25 /10 \approx 8 $ mm
Width at equator $ \pi 100/ 8 \approx 40$ mm

Draw sine curve as indicated, calculate pole to equator distance.  Prepare 8 of sine profiles starting 10 mm wide at north pole and 40 mm at equator, and laying them side by side and glue edges.
A: Scenario 2: The area is the same. (I think)
Label the corners of the square ABCD clockwise. Without loss of generality, place the square with corner A at the top of the sphere. Let corner C be azimuth zero in Or scaling up the sphere to the Earth and the metal square to the size of Ireland(-or-so): A is at the North Pole, C is in the direction of the prime meridian to London, B is in the direction of Moscow, and D is in the direction of Brasilia.  ​(Insert usual reference to the riddle about being somewhere on the Earth, walking south, west, north and ending up at the same point: where are you?. And note the science fiction story by Stephen Baxter in which the geometry of the Earth is not a scaled-up version of the geometry of an orange.)
For an intuitive appreciation that the area of the traced shape is the same, take a view that is face-on to the prime meridian. Form terraces along the parallels - imagine cutting into the side of a mountain to make terraces for agriculture, or that you're on stage looking out to an amphitheatre.
Cut the square of metal (or Ireland) into sections of annulus to cover the terraces.  For any set of terraces that are of finite width, the horizontal area of the terraces is approximately the same as that of the square. But the terraces are flat, and so the approximation is achieved without having the stretch the metal.  And as the width of the terraces approaches zero (handwavium here) the approximation becomes perfect - the areas become equal.
Alternatively, to understand that the 'roll-unroll' process is area preserving - rather than working with a square, work with a circle of radius $1/2$.  That way, we can just view the process edge-on and see that you're just tracing out a segment of the perimeter of a circle.
What does the traced object look like?  The (images of) AB and AD are easy - they follow the meridians (to Moscow and Brasilia).  The tricky ones are BC and DC.  We have that BC departs from (the image of) B along the parallel through B (at right angles to the meridian) because that's what the corner of the square does.  The same happens for DC.  So if you were to turn the sphere so that the great circle arc from B to C is directly in front of you, the image of BC would be inside the great circle arc.  Likewise for CD.
​
