Say that instead of four triangles along each edge we have $n$. First count the triangles that point up. This is easy to do if you count them by top vertex. Each vertex in the picture is the top of one triangle for every horizontal grid line below it. Thus, the topmost vertex, which has $n$ horizontal gridlines below it, is the top vertex of $n$ triangles; each of the two vertices in the next row down is the top vertex of $n-1$ triangles; and so on. This gives us a total of
$$\begin{align*}
\sum_{k=1}^nk(n+1-k)&=\frac12n(n+1)^2-\sum_{k=1}^nk^2\\
&=\frac12n(n+1)^2-\frac16n(n+1)(2n+1)\\
&=\frac16n(n+1)\Big(3(n+1)-(2n+1)\Big)\\
&=\frac16n(n+1)(n+2)\\
&=\binom{n+2}3
\end{align*}$$
upward-pointing triangles.
The downward-pointing triangles can be counted by their by their bottom vertices, but it’s a bit messier. First, each vertex not on the left or right edge of the figure is the bottom vertex of a triangle of height $1$, and there are $$\sum_{k=1}^{n-1}=\binom{n}2$$ of them. Each vertex that is not on the left or right edge or on the slant grid lines adjacent to those edges is the bottom vertex of a triangle of height $2$, and there are
$$\sum_{k=1}^{n-3}k=\binom{n-2}2$$ of them. In general each vertex that is not on the left or right edge or on one of the $h-1$ slant grid lines nearest each of those edges is the bottom vertex of a triangle of height $h$, and there are
$$\sum_{k=1}^{n+1-2h}k=\binom{n+2-2h}2$$ of them.
Algebra beyond this point corrected.
The total number of downward-pointing triangles is therefore
$$\begin{align*}
\sum_{h\ge 1}\binom{n+2-2h}2&=\sum_{k=0}^{\lfloor n/2\rfloor-1}\binom{n-2k}2\\
&=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}(n-2k)(n-2k-1)\\
&=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}\left(n^2-4kn+4k^2-n+2k\right)\\
&=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+2\sum_{k=0}^{\lfloor n/2\rfloor-1}k^2-(2n-1)\sum_{k=0}^{\lfloor n/2\rfloor-1}k\\
&=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+\frac13\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\left(2\left\lfloor\frac{n}2\right\rfloor-1\right)\\
&\qquad\qquad-\frac12(2n-1)\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\;.
\end{align*}$$
Set $\displaystyle m=\left\lfloor\frac{n}2\right\rfloor$, and this becomes
$$\begin{align*}
&m\binom{n}2+\frac13m(m-1)(2m-1)-\frac12(2n-1)m(m-1)\\
&\qquad\qquad=m\binom{n}2+m(m-1)\left(\frac{2m-1}3-n+\frac12\right)\;.
\end{align*}$$
This simplifies to $$\frac1{24}n(n+2)(2n-1)$$ for even $n$ and to
$$\frac1{24}\left(n^2-1\right)(2n+3)$$ for odd $n$.
The final figure, then is
$$\binom{n+2}3+\begin{cases}
\frac1{24}n(n+2)(2n-1),&\text{if }n\text{ is even}\\\\
\frac1{24}\left(n^2-1\right)(2n+3),&\text{if }n\text{ is odd}\;.
\end{cases}$$
It does not follow that there are then $\frac{8!}{(8-2)!\cdot 2!}$ ways. It is essentially just a non-sequitur. When you subtract $10$, it doesn't mean anything.
Let me provide an argument as similar as possible to your argument.
Pick a vertex. There are $7$ other vertices we can draw to. Therefore, there are $7$ diagonals for each vertex. There are $10$ vertices total, so the total counted is $7 \cdot 10$. But for each diagonal, there are $2$ vertices that we picked. Therefore, we counted each diagonal twice! The answer is therefore
$$\frac{7 \cdot 10}{2} = 35$$
Now, let's look at what the solution intended. There are $10$ vertices. A segment is defined by its two endpoints. Therefore, there are
$$\binom{10}{2} = 45$$
segments, and $10$ of these are the sides of the decagon. Therefore there are $45 - 10 =35$ diagonals.
Best Answer
Assuming that you're fine with triangles whose sides are not, themselves, parts of the diagonals, that will be fine for a heptagon whose vertices are in general convex position: we just need to subtract off triangles with two points on a diagonal.
To do this as generally as possible: if we fix a vertex $P$, the number of degenerate triangles including that vertex is $2\cdot \binom{7}{5}$ (and $2 \cdot \binom{n}{5}$ in general). The reasoning:
We multiply this by $7$ (by $n$) for all the possible ways to choose $P$.
So you want to take your original answer of $7\cdot \binom{7}{4}\cdot \left(\binom{7}{4}-1\right)$ and subtract $2 \cdot 7\cdot \binom{7}{5}$ for the final answer.
Although you asked specifically for non-regular heptagons, it's worth pointing out that for a regular heptagon, the problem is much worse. The first concern is that there might be triple intersections of diagonals: there aren't any for the regular heptagon, but there are some for the regular $n$-gon with larger $n$. The other problem is degenerate triangles like $\triangle ABC$ in the diagram below:
It's not even immediately obvious that $A$, $B$, and $C$ are collinear (they are), let alone how to count all such triples without actually going through the diagram and dealing with every degenerate triangle one at a time.