Nonexistence of a simple group of order 420

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.

A15 does have subgroups of order 14 or 28. Actually an odd cycle is an even permutation, so 7-2-2 and 7-cycle are in A15. Without futher deduction that Sylow 2 group is isomophic to V4, the group of 28 could be <(1,2,3,4,5,6,7),(8,9,10,11)(12,13,14,15)>, i.e. A group generated by a 7-cycle and a disjoint 4-cycles. By contradiction assume that G is simple, |G|=4 x 3 x 5 x 7, so no subgroup of index less than 7, making n5=21, n7=15. Denote N7 be the normalizer of a sylow-7 group, so |N7|=28 and |N5|=20.

• We first prove that n3=70 and n2=35
• show that N2=A4 (N2 isomorphic to A4) and P2=V4 (Klein 4 group)
• There is an element of order 14 in N7 and an element of order 10 in N5.
• G acts transitively on all subgroups of order 2 by conjugation.
• There is a subgroup U of order 2, s.t. <P2, P5, P7> <= NG(U), whose index is less than 7.

By sylow's theorem, n3=10, 28 or 70, n2=7, 15, 21, 35 or 105. If n3=7 (or |N3|=60), since N3 is not simple group of order 60, n5=1. That means a P5 normalizes a P3, contradicting to |N5|=20. By the same method, we rule out |N2|=60.
If n3=10 (or |N3|=42), then there is a P7 commute with P3 in N3, contradict to |N7|=28.
Leaving only n3=70.
Next we rule out n2=15 (or N2=28). If it were, we first prove that N7=N2=C28, i.e. N7 and N2 are in the same conjugation class, then contradict by counting elements. If N7 is not isomorphic to C28, we can conjugate any N2 so that it contains the same P7 subgroup. The the unqiue sylow-2 subgroup Q2 of N2 normalizes P7 but distincts from P2 in N7, since N7 is not isomorphic to C28. So P7 is normal in <P7, P2, Q2>, contradicts to |N7|=28. So N7=N2=C28. The total elements of order 3 and 5 is 2x70+4x21=224, and the total unique elements of all conjugations of N2 (or N7) is greater than 196, overflowing |G|.
Now we rule out n2=21 (or N2=20). By the same way, we argue that N2 and N5 are in the same conjugation class and isomorphic to C20. We first find out U of order 2 commute with a P7, then find a P5 and P2 commute with U, so U=<P7, P5, P2>, whose order is greater than 140.
Next we rule out n2=105, leaving only n2=35 (or N2=12).
Now we show that N2=A4. If not, then it is isomorphic to an abelian group, contradicts to N3=6.
Since A4 act transitively on subgroups of order 2 in V4 (the order 4 subgroup in A4), and G act transitively on all P2, by conjugation, completing the proof that G acts transitively on all order 2 subgroups of G.
Since P2=V4, a homomorphism from V4 to Aut(P7) cannot be injective, so there is an element of order 2 compute with element of order 7 in N7. The same reason, There is an element of order 2 compute with element of order 5 in N5.
Now that G acts transitively on any subgroup U of order 2, we can choose a P5, P7, P2 normalizing U, making order of NG(U) bigger or equal to 70. QED.