Not to worry, you've explained your question fine. However, your lack of familiarity with vectors is probably making this a lot more complicated than it could be.
Anyway, to get to the point: the correct way to calculate the force in the one-dimensional case is
$$\sum F = F(\text{Earth|Obj}) + F(\text{Moon|Obj}) + F(\text{Sun|Obj})$$
It's not correct to assume that the object is part of the Earth in this problem, because you're calculating the force on the object itself. In fact, it's meaningless to add up forces that act on different things. So in your first (incorrect) calculation, where you add one force on the object to two other forces on the Earth, you're absolutely going to get a nonsensical result.
When you get to doing this in the 3D case, you do the same thing except with vectors:
$$\sum \vec F = \vec F(\text{Earth|Obj}) + \vec F(\text{Moon|Obj}) + \vec F(\text{Sun|Obj})$$
This just means that you need to use a copy of this equation for each component, $x$, $y$, and $z$.
$$\begin{align}\sum F_x &= F_x(\text{Earth|Obj}) + F_x(\text{Moon|Obj}) + F_x(\text{Sun|Obj}) \\
\sum F_y &= F_y(\text{Earth|Obj}) + F_y(\text{Moon|Obj}) + F_y(\text{Sun|Obj}) \\
\sum F_z &= F_z(\text{Earth|Obj}) + F_z(\text{Moon|Obj}) + F_z(\text{Sun|Obj})\end{align}$$
This you've done correctly. You just made a mistake in calculating the equivalent mass change: note that $983.239\text{ N}$ is between the two extremes you calculated in the 1D case, so the equivalent mass it gives should be less than $59\text{ g}$ different from $100\text{ kg}$. In fact it corresponds to a difference in equivalent mass of about $14\text{ g}$, not $262\text{ g}$.
Let me first remind you of (or perhaps introduce you to) a couple of aspects of quantum mechanics in general as a model for physical systems. It seems to me that many of your questions can be answered with a better understanding of these general aspects followed by an appeal to how spin systems emerge as a special case.
General remarks about quantum states and measurement.
The state of a quantum system is modeled as a unit-length element $|\psi\rangle$ of a complex Hilbert space $\mathcal H$, a special kind of vector space with an inner product. Every observable quantity (like momentum or spin) associated with such a system whose value one might want to measure is represented by a self-adjoint operator $O$ on that space. If one builds a device to measure such an observable, and if one uses that device to make a measurement of that observable on the system, then the machine will output an eigenvalue $\lambda$ of that observable. Moreover, if the system is in a state $|\psi\rangle$, then the probability that the result of measuring that quantity will be the eigenvalue of the observable is
\begin{align}
p(\lambda) = |\langle \lambda|\psi\rangle|^2
\end{align}
where $|\lambda\rangle$ is the normalized eigenvector corresponding to the eigenvalue $\lambda$.
Specialization to spin systems.
Suppose, now, that the system we are considering consists of the spin of a particle. The Hilbert space that models the spin state of a system with spin $s$ is a $2s+1$ dimensional Hilbert space. Elements of this vector space are often called "spinors," but don't let this distract you, they are just like any other vector in a Hilbert space whose job it is to model the quantum state of the system.
The primary observables whose measurement one usually discusses for spin systems are the cartesian components of the spin of the system. In other words, there are three self-adjoint operators conventionally called $S_x, S_y, S_z$ whose eigenvalues are the possible values one might get if one measures one of these components of the system's spin. The spectrum (set of eigenvalues) of each of these operators is the same. For a system of spin $s$, each of their spectra consists of the following values:
\begin{align}
\sigma(S_i) = \{m_i\hbar\,|\, m_i=-s,-s+1,\dots, s-1,s\}
\end{align}
where in my notation $i=x,y,z$. So for example, if you build a machine to measure the $z$ component of the spin of a spin-$1$ system, then the machine will yield one of the values in the set $\{-\hbar, 0, \hbar\}$ every time. Corresponding to each of these eigenvalues, each spin component operator has a normalized eigenvector $|S_i, m_i\rangle$. As indicated by the general remarks above, if the state of the system is $|\psi\rangle$, and one wants to know the probability that the measurement of the spin component $S_i$ will yield a certain value $m_i\hbar$, then one simply computes
\begin{align}
|\langle S_i, m_i |\psi\rangle|^2.
\end{align}
For example, if the system has spin-$1$, and if one wants to know the probability that a measurement of $S_y$ will yield the eigenvalue $-\hbar$, then one computes
\begin{align}
|\langle S_y, -1|\psi\rangle|^2
\end{align}
Spinors.
In the above context, spinors are simply the matrix representations of states of a particular spin system in a certain ordered basis, and the Pauli spin matrices are, up to a normalization, the matrix representations of the spin component operators in that basis specifically for a system with spin-$1/2$. Matrix representations often facilitate computation and conceptual understanding which is why we use them.
More explicitly, suppose that one considers a spin-$1/2$ system, and one chooses to represent states and observables in the basis $B =(|S_z, -1/2\rangle, |S_z, 1/2\rangle)$ consisting of the normalized eigenvectors of the $z$ component of spin, then one would find the following matrix representations in that basis
\begin{align}
[S_x]_B &= \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \frac{\hbar}{2}\sigma_x\\
[S_y]_B &= \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \frac{\hbar}{2}\sigma_y\\
[S_z]_B &= \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} =\frac{\hbar}{2}\sigma_z\\
\end{align}
Notice that these representations are precisely the Pauli matrices up to the extra $\hbar/2$ factor. Moreover, each state of the system would be represented by a $2\times 1$ matrix, or "spinor"
\begin{align}
[|\psi\rangle]_B = \begin{pmatrix} a \\ b\end{pmatrix}.
\end{align}
And one could use these representations to carry out the computations referred to above.
Best Answer
Forget about the motion for a moment. First just consider a pendulum of length $\ell$ at an angle $\vartheta$ from rest. If $(0,0)=0\mathbf{e}_x+0\mathbf{e}_y$ is where the pendulum is attached, then the end of it is at $$ \bigl(\ell\cdot\sin(\vartheta), -\ell\cdot\cos(\vartheta)\bigr) $$
So, now let's describe the motion in terms of these variables. $\ell$ would usually be constant, so we don't need to care about this. It's $\vartheta$ that varies in time, so we get $$ y(t) = -\ell\cdot\cos(\vartheta(t)), $$ $$ x(t) = \ell\cdot\sin(\vartheta(t)), $$ Unfortunatly, we don't know $\vartheta(t)$ yet. We can either
Lets stick to small angles here. $\vartheta(t)=A\cdot\sin(\omega t)$ with $A\ll1$. We can then try to further simplify the cartesian coordinate representations: in $$ x(t) = \ell\cdot\sin(\vartheta(t)) = \ell\cdot\sin\bigl(A\cdot\sin(\omega t)\bigr), $$ we can use that $A\cdot\sin(\omega t)\ll 1$ and $\sin(x)\approx x$ for $x\ll 1$, therefore $$ x(t) \approx \ell\cdot A\cdot\sin(\omega t). $$ In $$ y(t) = -\ell\cdot\cos(\vartheta(t)) = -\ell\cdot\cos\bigl(A\cdot\sin(\omega t)\bigr), $$ we can use that $\cos(x)\approx 1-\tfrac{x^2}2$ for $x\ll 1$, therefore $$ y(t) \approx \ell\cdot \bigl(\tfrac{A^2}2\cdot\sin^2(\omega t)-1\bigr) $$ which, due to $\sin^2(x)=\tfrac12-\tfrac12\cos(2x)$, can be further simplyfied to $$ y(t) \approx \tfrac{A^2}{4}\ell\cdot\cos(2\omega t) - \tfrac\ell2. $$ As you see, two harmonic oscillators is in fact not such a bad assumption, and the important thing is that $y$ must oscillate at twice the frequency of $x$.
Sorry, I forgot about this question...
When I say $\sin(x)$ is approximately $x$, I mean just what I say: if you look at the sine function, and "zoom in" very close to 0 (that is, $x\ll1$, for instance $x=.001$), then this function looks very much like the simple $f(x)=x$. http://www.wolframalpha.com/input/?i=plot+sin%28x%29+from+-.01+to+.01 It's just when you look very carefully that you notice it's in fact bent a little bit.