Here's how I understood this before I studied calculus:
Suppose the projectile is to be launched from the point $(0,0)$
at initial speed $v_0$ in a direction making an angle $\theta$
above the positive $x$ axis,
and suppose (for now) that there is no gravity here.
Then the projectile travels a distance $v_0$ in that direction in the first
unit of time,
and a distance $v_0 t$ in that direction in the first $t$
units of time.
Construct a right triangle with legs parallel to the $x$ and $y$ axes
and hypotenuse $v_0 t.$
The horizontal leg is then $v_0 t\cos\theta$
and the vertical leg is $v_0 t\sin\theta.$
If we construct this triangle on the coordinate plane with the
hypotenuse in the initial direction of the projectile
with one end of the hypotenuse at $(0,0),$
we find that the $(x,y)$ coordinates at time $t$ are given by
\begin{align}
x &= v_0 t\cos\theta,\\
y &= v_0 t\sin\theta & \text{(in zero gravity).}
\end{align}
Now we consider the effects of gravity.
Assuming you have already derived the fact that an object starting
at $(0,0)$ with no initial velocity will fall under the influence of
gravity so that it reaches the vertical coordinate $y = -\frac12 g t^2$
at time $t,$
we combine the motion due to the initial velocity of the projectile
with the motion due to gravity.
The result is that the projectile ends up at a position displaced
vertically by $-\frac12 g t^2$ from where it would have been if there
were no gravity, and not displaced horizontally (because things don't fall sideways!), so it ends up at
\begin{align}
x &= v_0 t\cos\theta, \tag1 \\
y &= v_0 t\sin\theta - \tfrac12 g t^2 & \text{(where gravity is $g$).}\tag2
\end{align}
To really understand this, I think you need some notion of Galilean relativity: the launch speed and angle establish an inertial frame of reference traveling at velocity $v_x=v_0\cos\theta,$
$v_y=v_0\sin\theta$ relative to the ground, and the projectile falls
straight down within that frame of reference.
One way to explain it might be, you fire two cannonballs from the
same place in the same direction at the same speed; one is a magic cannonball that isn't affected by gravity, but the other is an ordinary cannonball.
A hypothetical observer riding on the magic cannonball will initially see the
ground receding down and behind while the ordinary cannonball is stationary,
but after $t=0$ the ordinary cannonball will appear to fall straight down
(as seen from the vantage point on the magic cannonball).
So while the magic cannonball follows a trajectory with $y=v_0t\sin\theta,$
the ordinary cannonball follows Equation $(2).$
If you launch the projectile from a point $(x_0,y_0)$ instead of
$(0,0),$ everything is displaced by $x_0$ horizontally and $y_0$ vertically,
and you get the more general equations
\begin{align}
x &= v_0 t\cos\theta + x_0,\\
y &= v_0 t\sin\theta - \tfrac12 g t^2 + y_0.
\end{align}
But as long as you launch from $(0,0),$ then $x_0 = y_0 = 0$ and the
equations end up acting like $(1)$ and $(2).$
Now observe that if $x = v_0 t\cos\theta$ (from Equation $(1)$), then
$$t = \frac{x}{v_0\cos\theta}.$$
Make this substitution for $t$ in Equation $(2)$ and you can work out
the equation for $y$ as a function of $x.$
I don't agree with the way the question implies that
parabolas of the form $y = ax + bx^2$
merely appear to be like trajectories in shape,
nor the implication that the formula with $v_0$ and $\theta$
is more accurate than $y = ax + bx^2.$
On the contrary, I would say that $y = ax + bx^2$ is just as
accurate an equation of a projectile trajectory (provided that $b$ is negative), even though it does not explicitly tell you what the initial velocity and angle are.
If you want the initial velocity and angle of the projectile that follows
the trajectory $y = ax + bx^2,$
you have to solve for $v_0$ and $\theta$ in terms of $a,$ $b,$
and $g.$
Update: For reference, it may be helpful to show the derivation of $v_0$ and $\theta$ from $a$ and $b.$ We suppose that we are given the path
$$ y = ax + bx^2 $$
for some given (known) values of $a$ and $b.$
This can only be a projectile trajectory if the same path is also given by an equation of the form
$$ y = (\tan\theta)x - \left(\frac{g}{2v_0^2\cos^2\theta}\right)x^2$$
for some values of $v_0,$ $\theta,$ and $g.$
In order for both equations to describe the same path, the right hand side must be the same polynomial in each equation. This means the coefficient of $x$ must be the same each time, and the coefficient of $x^2$ must be the same. So we have a projectile trajectory only if we can simultaneously solve these two equations:
\begin{align}
a &= \tan\theta, \\
b &= -\frac{g}{2v_0^2\cos^2\theta} = -\frac{g}{2v_0^2}\sec^2\theta.
\end{align}
From the first of these equations we have $\theta = \tan^{-1}(a)$
(for a projectile that travels over the part of the parabola for
which $x \geq 0$)
or $\theta = \tan^{-1}(a) + \pi$ (for the other part of the parabola).
Using the identity $\sec^2\theta = 1 + \tan^2\theta,$ we have
$$
b = -\frac{g}{2v_0^2}\sec^2\theta = -\frac{g}{2v_0^2}(1 + \tan^2\theta),
$$
and using the fact that $a = \tan\theta,$
$$
b = -\frac{g}{2v_0^2}(1 +a^2).
$$
If $g$ is known, then there is only one unknown in this last equation,
$v_0.$ To find $v_0$ we first isolate $v_0^2$ on one side of the equation:
$$
v_0^2 = -\frac{g(1 + a^2)}{2b}.
$$
The left side must be positive; making the usual assumption that $g > 0,$ and and noticing that $1 + a^2$ is necessarily positive,
we conclude that we can only solve this equation if $b < 0.$
Provided that $b < 0,$ we find that
$$
v_0 = \sqrt{-\frac{g(1 + a^2)}{2b}},
$$
so we have derived both the angle and initial velocity from the coefficients $a$ and $b.$ Moreover, this derivation always works as long as $b < 0$:
any downward-opening parabola is the path of a projectile trajectory.
