this is how i would explain your approximation.
First start with notation:
Define $K_{atm}$ to be the atm strike.
Define $\Delta K := K2 - K1$ where $K2 > K_{atm} > K1$. This corresponds to $\Delta K = $$StrD$ in your notation.
Now assume a black scholes world, within this world we can approximate the Call and Put price of an atm option with:
$C_{atm} = P_{atm} = 0.4 \sigma \sqrt{T} F$.Therefore the price of a straddle is given by $0.8 \sigma \sqrt{T} F$.
The delta of an option is given by $N(d1)$. For $K_{atm}$ we get that $d1 =\frac{ 0.5\sigma^2 T}{\sigma \sqrt{T}}$. So $N(d1)$ is close to 0.54. Now we evaluate $N(d1)$ at $K = K_{atm}+-straddle$. For example suppose r = 0, S = 100, T = 1, vol = 0.2 then K_{atm} = 100, the straddle = 0.8*0.2*1*100 = 16 (and a vanilla call is priced at 8). Question: what is the delta when evaluated at K=116 or 84? Answer: delta is 0.26 and 0.84. Comparing 0.26 against 0.54 and 0.84 against 0.54 we see that for the up and down straddle move the delta changes about 0.3. Furthermore, by simply plotting the delta against strikes we can observe that the delta is approximately linear between the these strike levels.
Therefore we can explain your $\Delta CS$.
You have a move of 0.3 deltas for 1 move of size straddle. thefore a move of size $\delta K = (K_{new} - K_{atm})$ changes the delta by: $0.3 \frac{\delta K}{straddle}$. Alternatively we can write the Delta as function of K. $\Delta(K) = 0.50 + \frac{0.3(K - K_{atm})}{straddle}$
The price of a callspread is given by $C(K1) - C(K2)$. Now we approx C(K2) by:
$C(K2) = C(K1) + \int_{K1}^{K2} \Delta(K) dK$. The value of your callspread therefore is: $\int_{K1}^{K2} \Delta(K) dK$.
I think integrating the above gets you very close to the formula that you want.
