How can {→,⊕} be complete when {¬,⊕} isn't? Please provide an example with {→,⊕} that can't be realized with {¬,⊕}. I can't think of what I can't realize with  {¬,⊕}. 
I have a simple model where I think the bottom is more like YES ⊕ NO (and it doesn't at all look like YES OR NO even though the picture doesn't say that YES and NO can't both be true for the same input(?)

It depends what we want the algorithm to do but usually the output should be neither "both YES and NO" nor "neither YES nor NO". 
I understand we might want the cases with no answer for one case of input or both yes and no answer for another case of input. 
The system I was modelling would output 0 XOR 1 and never something else. 
 A: Show that $\lor$ cannot be produced using only $\neg$ and $\oplus$. One way to do this is to show that if $f(p,q)$ is any proposition in two variables constructed solely with $\neg$ and $\oplus$, the $4$-row truth table of $f$ has a $\mathsf{T}$ in an even number of rows. You can do this by structural induction on $f$: show that if $f$ has the property, then so does $\neg f$, and that if $f$ and $g$ have the property, then so does $f\oplus g$.
Note that you can replace $\lor$ by any connective that has an odd number of $\mathsf{T}$’s in its truth table; $\land$ works equally well, as does $\to$.
A: It looks like you're confused about what "complete set of connectives" means.
By definition, what the claim "$\neg$ and $\oplus$ form a complete set of connectives" means is:

Every possible truth table is the truth table for the output of some expression built with only $\neg$ and $\oplus$ in addition to the input variables (and parentheses etc.)

Here, of course "possible truth table" implies that the ground rules for truth tables are followed -- every combination of input values must appear in exactly one line, but the output values are arbitrary, that is, no matter which output values we choose it must be possible to come up with an expression that produces exactly these outputs.
Since the claim we're investigating is one about "every truth table", it willl be false (and therefore $\{{\neg},{\oplus}\}$ is not a complete set) if we can find even a single truth table that doesn't have a matching expression. It is easy for an "every X" statement to go wrong, hard for it to hold.
Here's a truth table that has no matching expression built with only $\neg$ and $\oplus$:
 input 1     input 2   |   output
 --------------------------------
    0           0      |     0
    0           1      |     1
    1           0      |     1
    1           1      |     1

(It happens to be the truth table for $\lor$, but that is not important, except possibly as a way to name the truth table without writing it down explicitly. What is important here is that we're interested in whether $\{\neg,\oplus\}$ can make every truth table. Here's some random truth table -- can we make it or not?)
It turns out that we can't. It is quite possible to find an expression with two variables (and all connectives $\neg$ or $\oplus$) such that inputs 0,0 produce output 0, and a different expression such that inputs 0,1 produce output 1, and so forth. But there's no single expression that produces the right answer for every line of the truth table.
(For proof that this particular truth table cannot be made, see the answer I linked to earlier).
So we know that this particular truth table cannot be generated with $\neg$ and $\oplus$ alone. We can then stop worrying about which other truth tables are possible or not; the existence of even one that can't be produced means by definition that $\neg$ and $\oplus$ is not complete.
A: There are two ways to look at boolean connectives/functions. The usual way it to let $\mathbf B = \{0,1\}$, so you may think of them as functions $\mathbf B^i \to \mathbf B$. This is the classical interpretation with a truth table. For instance, $\land(0,1) = 0$.
But alternatively, you may think about boolean functions as polynomials over the field $\mathbf F_2 = \{0,1\}$. Then the 'addition' matches $\oplus$ and 'multiplication' matches $\land$. Now note that in this language
$$ \begin{aligned} \lnot x &= 1+x \\ x\land y &= xy \\ x\lor y &= \lnot(\lnot x \land \lnot y) = 1 + (1+x)(1+y) = x+y+xy \\ x \oplus y &= x+y\\ x\to y &= (\lnot x \lor y) = (1+x)\lor y = 1+x + y + y + xy = 1+x+xy\dots \end{aligned} $$
This means that if you are only allowed to combine $\oplus$ and $\lnot$, then the only functions you can make by composing these are the affine ones, this mean functions of the form
$$ a_0 + a_1 x_1 + a_2 x_2 + \dots + a_nx_n. $$
Where $a_i \in \{0,1\}$. (In particular, for $2$ variables $x$ and $y$, there are $8$ affine functions namely $0$, $1$, $x$, $1+x$, $y$, $1+y$, $x+y$ and $1+x+y$.)
However, $\to$ is clearly non-affine, because it contains the second degree term $xy$, which shows that it cannot be created from composition of $\oplus$ and $\lnot$.
