Consider the algebra $A_n$ generated by $u_1,\ldots,u_{n-1}$ subject to relations $u_i^2=-2u_i$, $u_iu_{j}u_i=u_i$ for $|i-j|=1$ and $u_iu_j=u_ju_i$ for $|i-j| \geqslant 2$. The algebra $A_n$ is the Temperley-Lieb algebra. This algebra has a basis consisting of reduced words $u_I=u_{i_1}\ldots u_{i_k}$.
There is a surjective homomorphism $\varphi\colon \mathbb{C}[S_n] \twoheadrightarrow A_n$ given by $(ii+1) \mapsto u_i+1$. The question is the following: is there any nice description of $\varphi(w)$ for $w \in S_n$? Is it true that $\varphi(w)$ is the linear combination of different $u_I$ with coefficients $\pm 1$?
What you're describing has to do with the $n$th tensor power of the $2$-dimensional standard representation $V_1$ of $\mathfrak{sl}(2,\mathbb{C})$ along with the commuting action by $S_n$. More precisely, the images of $U(\mathfrak{sl}(2,\mathbb{C}))$ and $\mathbb{C}[S_n]$ in the algebra $\operatorname{Hom}(V_1\otimes V_1\otimes\dots\otimes V_1)$ where there are $n$ factors in the tensor product, are the subalgebras involved in Schur-Weyl duality. The Temperley-Lieb algebra $A_n$ (with loop value $-2$, as in your presentation of it) describes the subalgebra corresponding to the image of $\mathbb{C}[S_n]$.
Let's spend some time going through how this works. First of all, let $V_1=\mathbb{C}^2$, where $\mathfrak{sl}(2,\mathbb{C})$, thought of as traceless $2\times 2$ matrices, acts on $V_1$ by the usual matrix-vector multiplication. We'll let $\{e_1,e_2\}$ be the standard basis for $V_1$. There is an invariant nondegenerate bilinear form $\beta:V_1\otimes V_1\to \mathbb{C}$ given by $$ \beta(v\otimes w)= v^T\begin{bmatrix}0&-1\\1&0\end{bmatrix}w $$ (and in fact, all other such forms are a scalar multiple of this one). With respect to the basis, we can write this as $\beta=e_2^*\otimes e_1^*-e_1^*\otimes e_2^*$. There is a sort of "inverse" $\alpha:\mathbb{C}\to V_1\otimes V_1$ to this defined by $$\alpha(1)=-e_2\otimes e_1+e_1\otimes e_2.$$ I'm more comfortable with explaining the following with some graphical tensor notation, but I'll also give some algebraic formulae in parallel. We'll represent $\beta$ and $\alpha$ by
These satisfy the following identities:
We can write the first as $(\beta\otimes\mathrm{id})\circ(\mathrm{id}\otimes\alpha)=\mathrm{id}$, the second as $(\mathrm{id}\otimes\beta)\circ(\alpha\otimes\mathrm{id})=\mathrm{id}$, and the third as $\beta\circ\alpha=-2$.
By the way, where $\beta$ comes from is an isomorphism between $V_1$ and its dual representation $V_1^*$, and then $\alpha$ comes from the inverse of the matrix for $\beta$.
There's a convention for the Temperley-Lieb algebra (I believe due to Kauffman) where you draw the above diagrams as curves with isolated critical points, where each critical point is interpreted as $\alpha$ or $\beta$ depending on the sign of the second derivative. With this convention, the above diagrams look like
This lets the diagrams be more concise while letting our topological intuition help us perform calculations: you can "pull strings straight"!
With some of the background out of the way, what I want to do is write the swap homomorphism $\tau:V_1\otimes V_1 \to V_1\otimes V_2$ defined by $\tau(v\otimes w)=w\otimes v$ in terms of $\alpha$ and $\beta$. Writing $\tau$ as a "crossing" in a diagram, we can verify the following identity holds:
Algebraically, this is that $\tau=\mathrm{id}+\alpha\circ\beta$.
Now let's spell out the correspondence between $A_n$ and the above so-called string diagrams. We set
and define a multiplication operation by vertical stacking. Hence, $u_i^2=-2u_i$ since that forms a closed loop in the diagram, and $\beta\circ\alpha=-2$:
There is an obvious homomorphism $\mathbb{C}[S_n]\to \operatorname{Hom}(V_1^{\otimes n})$ given by sending a permutation to the induced permutation of the tensor factors. The above identity that $\tau=\mathrm{id}+\alpha\circ\beta$ says that this homomorphism is equivalently given by sending the transposition $(i\ i+1)$ to $\mathrm{id}+u_i$. Hence, the homomorphism surjects onto $A_n$.
It turns out that a basis for $A_n$ corresponds to all possible ways of connecting $n$ points to $n$ points with curves that don't cross (counted by the Catalan numbers). For example, here are the five diagrams that form a basis for $A_3$:
Algebraically, we can give these as $$\{1,u_1,u_2,u_1u_2u_1,u_2u_1u_2\}$$
As a partial answer to your first question, the graphical presentation of elements of $A_n$ gives a nice way to calculate the images of elements of $S_n$, since the graphical notation gives a canonical way to describe a basis: they are the elements where each strand has 0 or 1 critical points. The data of such an element is simply the pairs of endpoints for each strand. For example, an expansion of an element of $S_3$:
I'm not familiar with any nice formulas, though, for finding the coefficient of a particular basis element for a specific element of $S_n$.
The answer to your second question is no. In $S_4$, the image of $(1\ 3)(2\ 4)$ has a coefficient of $2$ in front of $u_2$. Here's an expansion of this permutation into its sixteen terms, with the ones evaluating to $u_2$ highlighted in yellow:
Since loops give a factor of $-2$, we see the coefficient of $u_2$ is $1+1-2+1+1=2$.
We could have avoided writing all sixteen diagrams by thinking about what choices need to be made at each crossing to achieve $u_2$. You know that $u_2$ involves certain boundary points being connected together in a certain pattern, so you could start by, for example, finding all paths that connect the bottom left corner to the upper left corner, then recursing for each strand of $u_2$.
(By the way, some useful notes by Scott Morrison about diagrams for the Temperley-Lieb algebra.)