Let (x^2)/(f(a^2+7a+3)) + (y^2)/(f(3a+15)) = 1 represent an ellipse...

For the given equation to represent an ellipse with its major axis along the $y$-axis, the denominator of $y^2$ must be strictly greater than the denominator of $x^2$.

$f(3a+15) > f(a^2+7a+3)$

Since $f$ is given as a strictly decreasing positive function on $\mathbb{R}$, the inequality sign reverses for the arguments:

$3a+15 < a^2+7a+3$

$a^2 + 4a - 12 > 0$

$(a+6)(a-2) > 0$

The solution to this inequality is $a \in (-\infty, -6) \cup (2, \infty)$.

This can be rewritten in terms of the set difference as $a \in \mathbb{R} - [-6, 2]$.

Comparing this with the given set $\mathbb{R} - [\alpha, \beta]$, we get $\alpha = -6$ and $\beta = 2$.

$\alpha^2 + \beta^2 = (-6)^2 + (2)^2 = 36 + 4 = 40$

Answer: $40$