Note: Although the question looks like a physics question, the main problem that I'm having is about the math and with the geometry.
We know that in order for a spherical mirror to focus, the equation $$y^2 = 4fx - x^2$$ needs to behave like the parabola equation $$y^2 = 4fx,$$ so we need $$f>> x.$$
However, it is told me that the inequality $u >> h_1$, and similarly $v>h_2$ should also hold, but why is that so ?
I mean how can we show that $f >> x$ implies $u >> h_1$.
Edit: By Fermat's principle, we have $$\theta = \phi,$$$ and the lengths of both rays (actually any ray reflected by the mirror going between the source and the dest. ) are equal.
This is a bit less of a brute force approach, a bit more geometry, with a lot of trig and some calculus involved. Since I want to demonstrate that the $h \ll u$ is a condition independent from $x \ll f$, we will consider a picture like this:
$A$ is the source and the rays reflect so that the line from the point of incidence $P$ to the center $C$ is a bisector. I have fixed $u = 2f$ (in other words $AC$ is perpendicular to the main axis $OC$) because this greatly simplifies calculations. In any case, since I want to show that the $h \ll u$ condition is necessary, it should be enough to show that it is necessary in this particular case.
All the different rays that reflect from different positions of $P$, in general, aren't concurrent. We will consider $AO$ as a sort-of "central" axis for the (not perfectly focused) image. As point $P$ moves around the spherical surface, the image $B$ also moves along this axis. Our aim is to quantify the "spread" of those positions. Since the ratio $h/2f$ will feature prominently in our calculations, let's call it simply $\eta = h/2f$.
Parametrize the position of point $P$ through the angle $\phi$. Then out first step is to find $\theta$ as a function of $\phi$. In $\Delta APC$, $\angle PCA = \pi/2 - \phi$, hence $\angle PAC = \pi/2 -(\theta -\phi)$. Apply the sine rule $h/\sin\theta = 2f/\cos(\theta - \phi)$. Cross-multiply, expand the cosine, then solve for $\tan\theta$: $$ \tan\theta = \frac{\eta\cos\phi}{1-\eta\sin\phi}. $$
Now let $OB = s(\phi)$. To find it, first use the isosceles $\Delta OPC$ to find the segment (not arc) $OP = 4f\sin(\phi/2) = 2f \sin\phi/\cos(\phi/2)$. Then from $\Delta OPB$, in which some angle manipulations produce $\angle OPB = \angle OPC - \theta = \pi/2 - \phi/2 -\theta = \pi/2 - (\theta + \phi/2)$ and similarly for $\angle POB$, so that $\angle OBP = \pi - \angle OPB - \angle POB = \phi + \theta - \theta_0$, again the sine rule tells us $$ \frac{s}{\cos(\theta + \phi/2)} = \frac{OP}{\sin(\phi + \theta - \theta_0)} $$ from where $$ s = \frac{2f\sin\phi\cos(\theta + \phi/2)}{\cos(\phi/2)\sin(\phi + \theta - \theta_0)}. $$ Notice that in this formula, in addition to the explicit dependence on $\phi$ there is also implicit through $\theta(\phi)$. Some trig identities can help us make it explicit. The idea is to expand all the trig functions containing more than one term, then use the expression we have for $\tan\theta$ to replace every occurrence of $\sin\theta$ with $\cos\theta$ (times the expression for $\tan\theta$ as a function of $\phi$), eventually all the $\cos\theta$'s get factored out and cancelled. Similarly with $\theta_0$ using $\tan\theta_0 = h/2f = \eta$, but this time one $\cos\theta_0$ survives in the denominator: $$ s(\phi) = \frac{2f}{\cos\theta_0} \cdot\frac{\sin\phi}{\cos(\phi/2)}\cdot\frac {\cos{\phi\over 2} - \frac{\eta\cos\phi}{1-\eta\sin\phi}\sin{\phi\over 2}} {\sin\phi\left(1 + \frac{\eta^2\cos\phi}{1-\eta\sin\phi}\right) + \cos\phi\left(\frac{\eta\cos\phi}{1-\eta\sin\phi} - \eta\right)} $$
Now get rid of denominators in the big fraction, and after some applications of double angle and half-angle identities, you can get $$ s(\phi) = \frac{2f}{\cos\theta_0} \cdot \frac {\sin\phi + \eta(\cos 2\phi - \cos\phi)} {\sin\phi + \eta(\cos 2\phi - \cos\phi) + \eta^2\sin 2\phi}. $$
This is already pretty nice, though is you want to compute the Taylor expansion by hand, it is worth playing around with it some more. Anyway, everything so far has been without any approximations. You could plot it for some values of $\eta$ to get a sense for how it behaves.
Now suppose that the mirror is cut so that $\phi$ takes values in the interval $[-\phi_{max}, \phi_{max}]$. The measure of "blurriness" is the corresponding range of values of $s$. Suppose $\phi_{max} \ll 1$ (this is to say, we only have a small portion of the sphere, and is roughly equivalent to your $x \ll f$). Is it enough to get a sharp image? For small $\phi$'s, we can look at the Taylor series expansion of $s(\phi)$ around $0$: $$ s(\phi) = \frac{2f}{\cos\theta_0} \cdot \left\{ \frac{1}{1+2\eta^2} -\frac{3\eta^3}{(1+2\eta^2)^2}\phi + \frac{\eta^2(2-5\eta^2)}{(1+2\eta^2)^3}\frac{\phi^2}{2} + \dots \right\} $$
So the first two terms after the constant (which is of order $~1$) are of order $~\eta^3\phi$ and $\eta^2\phi^2$. If $\eta$ is not small, the first of those will be more significant, and we see how sensitive it is to the value of $\eta$: points closer to the axis will have very much sharper images. Sure, we can still get a good focus by making $\phi_{max}$ very small - jsut like a pinhole camera, actually - but this has disadvantages, such as the image will be very dim (not much light reflects off a small-area mirror). If we want a bright image, we don't want $\phi_{max}$ to be too small, and then for good focus we will want $\eta \ll 1$ as well. The two conditions are not equivalent, but rather they complement each other.
A couple of side notes:
My analysis was limited to the plane containing the axis of the mirror and the source point. In reality, a 3-d mirror will also reflect light rays that travel at an angle to the plane of the diagram, and those might not even intersect the line $OB$; in other words, point $B$ does not even exist because the lines we want to intersect are skew. In that case we would probably consider the shortest distance between these lines as a measure of "lateral blurriness" (the function $s(\phi)$ we derived was a description of blurriness "along" the axis).
Interestingly, the constant term in the Taylor expansion didn't turn out to be $1$, as we would have expected: for a source point $2f$ from the mirror, the image should be also at $2f$ (horizontally), i.e. $s_0 = 2f/\cos\theta_0$ - that's what the lens equation tells us. But we find that this only holds for $\eta \ll 1$. So even if the image is not too blurry, it is not where it should be, meaning the image of an extended object will be distorted. This is a fact that is all too familiar to anyone wearing eyeglasses: when not looking through the center of the lens, their vision is both less clear and distorted.