"the gravitational potential energy should DECREASE as it is converted into kinetic energy, but I guess I'm not really sure why it's negative at position 3."
To try and explain this I'll use an analogy. Imagine for a second you have a cat, and you throw the cat up into the air. If you choose to use the ground as a reference point, then the cat starts off with some vertical displacement, and will arc its way downwards until it hits zero, when it hits the ground. Instead, lets try taking the moment the cat leaves our hands as a reference point. Instead, the cat's distance away from you will increase for a while, come to a standstill, drop PAST your hands, and become negative. In a sense, that's the theory behind the statement you've made. Gravitational potential energy really doesn't have any meaning unless you state from which reference point your'e taking the measurement from, just as the cat's height has no meaning unless you say whether you are measuring from your hands or from your feet.
"but why is −mgxsinθ negative and why isn't kx2 negative?"
Because xsinθ is a distance, it will be positive. But that's not what we really want here. Since the block is BELOW the reference point we've taken (point 2), its distance is actually negative. Take the cat example I used, if we take our hands as the reference point, and the cat is below them, it follows that the distance in the upward direction is negative.
kx^2 isn't negative, because the distance in this case is assumed to be inwards toward the spring, as it represents the elastic potential energy. Let's take an analogy. Say the cat that we have thrown arcs, and instead of landing on the floor lands on a trampoline. Logically, the more that the trampoline bends inward, the greater force the cat will feel, and the greater potential the cat will have to fly off into space. This means that the cat's Elastic potential energy will have some correlation to the distance travelled into the trampoline. It's the same principle for the spring, the greater value of x, the more you travel into the spring, the more potential energy you have.
"And finally, how come one side is equal to the other if the right side involves a completely different aspect to it (the spring) that isn't found in the initial conditions. "
Oh, but it IS found in the initial conditions. Here's what the full equation would look like.
Gravitational Potential energy + Kinetic energy + Elastic potential energy (reference point 1)
= Gravitational Potential energy + Kinetic energy + Elastic potential energy(Reference point 3)
But then where did it go on the left hand side? Simple, the block was just not touching the spring yet, so the value of the Elastic potential energy at point one was ZERO, (similar to how the value of the kinetic energy at point 3 is zero).
Hope I answered your question, if something needs clarifying reply, it's my first answer here so it might not be very clear.
How do you lift a body up from the ground?
The body remains at equilibrium under the acting of opposite forces on it: the $\uparrow$ normal force from the ground & the $\downarrow$ gravitational force . In order to move it up, you have to give some infinitesimal $\uparrow$ force greater than gravitational force which would accelerate the body upwards.
Let the force be $\mathbf F_\text{ext}.$ Therefore the equation of motion provides:
$$\mathbf F_\text{ext}- m\mathbf g = m\;\delta \mathbf a\implies \mathbf F_\text{ext}= m\mathbf g + m\; \delta \mathbf a \;.$$
Let the body gets displaced above by $\bf h\;.$ Therefore the work done by $\mathbf F_\text{ext}$ is given by \begin{align}W_{\mathbf F_\text{ext}}&= \mathbf F_\text{ext}\cdot \mathbf h\\ &= m\mathbf {g}\;\cdot \mathbf h + m\; \delta \mathbf a \;\cdot\mathbf h \;.\end{align}
This work can be assessed as $$\overbrace{m\mathbf {g}\; \cdot\mathbf h}^{\text{work done against gravity}} + \overbrace{m\; \delta \mathbf a \;\cdot\mathbf h}^{\text{amount of kinetic energy increased}}\;.$$
The former work is what you call potential energy- the work done against the gravitational force. Hadn't the force worked against the gravitational force, then the body would have been constantly decelerated by the gravitational force turning all the kinetic energy into potential energy of the system.
At the present context, the later work is zero; so the body is moving at constant velocity.
But how?
There must be a force $\mathbf F_\text{ext}$ to counteract the gravitational force; otherwise the body couldn't move at constant velocity as its kinetic energy would have been converted to potential energy of the system. To counteract the work done on the body by the gravitational force, same amount of work is done by the external force on the body so as to retain its constant velocity; that work turns up as gravitational potential energy.
So, although the net work $W_{\mathbf{g}} + W_{\mathbf F_\text{ext}}= -m\mathbf g\;\cdot \mathbf h+m\mathbf g\;\cdot \mathbf h= 0 \;,$ in order for the body to move at constant velocity, the external force must have to do the same amount of work as done by the gravitational force- this is the potential energy of the system.
Best Answer
The external force is doing work on the system. The work it does increases the energy of the system. If the spring weren't there, it would increase the kinetic energy of the mass. Because the spring is there, it is increasing the potential energy of the spring.
Looking at the work done only on the mass is not sufficient to account for all the work done on the system.
When we lift a mass into the air at constant speed, the same situation is happening. There is no net work done on the mass (speed is constant), but work is done on the mass/earth system to increase the energy of that system.
Don't think of the potential energy being stored in an object. Instead think of it as the property of your system (be it spring, gravitational, or whatever).
Equivalently you can look at the work being done by the forces. Although the net work done on the mass is zero, the external force is doing positive work and the spring force is doing negative work.
The spring force doing negative work on the mass means that the mass is doing positive work on the spring. This positive work on the spring is increasing the potential energy stored in the spring.