As an answer for you first questions: let me call the homomorphism $\phi:G\rightarrow A_{15}$, and the order 7 element $g$, generating the Sylow 7-subgroup $P$, which has normaliser (the stabilisers of your action are called normalisers) $N_G(P)$ with order 28. First, you state that $P$ is the unique order 7 subgroup of $N_G(P)$, which is true, because an order 28 group has only 1 Sylow 7-subrgroup. Note that $\phi(g)^7=\phi(g^7)=\phi(1)=\text{id}$, so $\phi(g)$ (which permutes 14 letters as you say) has order dividing 7, i.e. either 1 or 7. Therefore it is either the identity permutation, a single 7-cycle, or 2 disjoint 7-cycles. But in any case apart from the last case, $\phi(g)$ fixes at least 7 other letters, in other words it normalises some Sylow 7-subgroups that are not $P$; call one of these $Q$. But now $N(Q)$ contains 2 distinct order 7 subgroups, namely $P$ and $Q$, which contradicts your point that an order 28 group can have only 1 order 7 subgroup. Therefore we get the result that you want; $\phi(g)$ is 2 disjoint 7-cycles.
EDIT: As an answer to the main problem: I don't think I would go down your route directly (mainly because I don't like composing permutations, which I find very fiddly), so here is a way to continue with minimal fiddly permutation computations.
Let's look at the structure of $N_G(P)$; we know it has at least one order 4 (Sylow 2) subgroup, so call this $H$. now we have $P\unlhd N_G(P)$ and $H\le N_G(P)$, so $PH\le N_G(P)$. But $P,H\le PH$ so the size of $PH$ is divisible by 4 and 7, so it has size (at least) 28, and hence $PH=N_G(P)$; we can represent all elements of $N_G(P)$ as the product of an element of $P$ with one of $H$, and this representation is unique because there are only $7\cdot4=28$ possible representations. $P\cong C_7$, and either $H\cong C_4$ or $H\cong V_4$. Then we have (by $P$ normal in its normaliser) that $N_G(P)$ is isomorphic to a semidirect product, $C_7\rtimes C_4$ or $C_7\rtimes V_4$, determined by a homomorphism $\psi:H\rightarrow\text{Aut}(P)\cong\text{Aut}(C_7)\cong C_6$. $\ker(\psi)$ has size 1, 2 or 4. But ${H\over\ker(\psi)}\cong\text{im}(\psi)\le C_6$, so it cannot have size 1. If it has size 4, then the homomorphism is trivial, the semidirect product is direct, and $N_G(P)$ is abelian, and hence either $C_{14}\times C_2$ or $C_{28}$. But as you say, it cannot have an order 14 or order 28 element, contradiction. So $\vert\ker(\psi)\vert=2$, and so the image is the order 2 subgroup of $\text{Aut}(P)$, namely the trivial automorphism and the inversion automorphism (call this automorphism $\varphi$).
First, suppose $H\cong V_4$, and call the 3 non-identity elements $x$, $y$ and $xy$. If $\psi(x)=\psi(y)=\varphi$, then $\psi(xy)=\varphi^2=\text{id}$, so in any case we have non-identity element $h\in H$ such that $\psi(h)=\text{id}$, i.e. $h$ commutes with $g$. But then (easily verified) $gh$ has order 14, contradiction again.
So we must have $H\cong C_4$, so there exists $h\in H$ with order 4. If $\psi(h)=\text{id}$, then $\varphi$ is not in the image of $\psi$ ($h$ generates $H$), contradiction. So $\psi(h)=\varphi$.
EDIT 2: as pointed out in the comment, at this point I was already done, because $h^2$ now commutes with $g$, so as before we have an order 14 element $gh$ which is a contradiction. If you want to read a far more convoluted solution, here it is:
By semidirect product, $hgh^{-1}=g^{-1}$. Then finally we do have to look at permutations again; let $\phi(h)=\sigma$, which should be an order 4 permutation by the injection $\phi$. We have $\sigma\phi(g)\sigma^{-1}=\phi(g)^{-1}$, where w.l.o.g. $\phi(g)$, two disjoint 7-cycles, sends $i$ to $i+1$ (mod 7) and $i'$ to $i'+1'$ (mod 7') as a permutation of the letters 0 through 6 and 0' through 6'. Working modulo 7 and 7':
If $\sigma(i)=j$ then $\sigma(i+1)=[\sigma\phi(g)\sigma^{-1}](j)=[\phi(g)^{-1}](j)=j-1$, and by induction $\sigma(i+k)=j-k$ for all $k$ modulo 7. Let $k=j-i$, and we get $\sigma(j)=i$. Similarly, if $\sigma(i')=j'$, then $\sigma(j')=i'$.
By the same inductive step we get that if $\sigma(i)=j'$ then $\sigma(i+k)=j'+k'$ for each $k$. But now $\sigma(j')=i+n$ for some particular $n$, which by the same induction again gives $\sigma(j'+k')=i+n+k$ for each $k$. Now $\sigma(i+n)=j'+n'$, and then by having order 4, $i=\sigma^{4}(i)=\sigma^{3}(j')=\sigma^{2}(i+n)=\sigma(j'+n')=i+n+n=i+2n$ so in fact we must have $n=0$, by inverse of 2 existing modulo 7. Hence $\sigma(i)=j'$ gives $\sigma(j')=i$. Again similarly, we also get that if $\sigma(i')=j$ then $\sigma(j')=i$.
This covers all cases for where $\sigma$ sends the letters 0 through 6 and 0' through 6', so $\sigma$ is composed only of disjoint transpositions, and hence has order 2, the final contradiction we needed, therefore $G$ is not simple.
Best Answer
When we're talking about the degree of a permutation group, the actual group structure doesn't matter. So in your example, the fact that $5 \equiv 2 \pmod 3$ is completely irrelevant; all that matters is that $C_3$ is (somehow) acting on $\{1, 2, 3, 4, 5\}$, and consequently will be said to have degree $5 = \left\lvert \{1, 2, 3, 4, 5\} \right\rvert$ in this case. In fact, if we're thinking about the action of any group on $\{1, 2, 3, 4, 5\}$, it will be said to have degree $5$, in that situation.
The term degree in this context is relative; it's not an intrinsic property of a group, but a property of the group action. It's similar to how a polynomial like $x^2 + 1$ would be considered irreducible -- over $\Bbb R$. But it can be factored as $(x - i)(x + i)$ over $\Bbb C$; the context is extremely relevant.
So, it's probably helpful to think of a permutation group as a whole bunch of data tied together: a group $G$, a set $\Omega$, and the action $G \times \Omega \to \Omega$ (or if you prefer, the homomorphism $\phi: G \to {\rm Sym}(\Omega)$) of $G$ on $\Omega$.
Example: The group of rotational symmetries of a cube is the symmetric group ${\rm Sym}(4)$, the full permutation group of the four diagonals of the cube. Thus we can think of ${\rm Sym}(4)$ as a permutation group in several ways, including but definitely not limited to:
A permutation group of degree $4$, if we think of it as acting on the diagonals of the cube.
A permutation group of degree $6$, if we think of it as acting on the faces of the cube.
A permutation group of degree $12$, if we think of it as acting on the edges of the cube, or
A permutation group of degree $8$, if we think of it as acting on the vertices of the cube.
Any subgroups of ${\rm Sym}(4)$, including $C_3$, also act on any of the above sets, and would have the same degrees in each situation.