write the trasition map $\phi$ between $\sigma_1$ and $\sigma_2$. Verify $\det( J(\phi))$ and find $T_p(S)$. Sphere $$S=\{(x,y,z) \mid x^2+y^2+z^2=R^2\}$$
$$
\sigma_1(u,v)=(u,v, \sqrt{1-u^2-v^2}) \\
\sigma_2(u,v)=(\tilde u, \sqrt{1-\tilde{u}^2 -\tilde{v}^2}, \tilde v)
$$
I guess $\{\sigma_1, \sigma_2\}$ forms an atlas for $S$. 
Question: write the trasition map $\phi$ between $\sigma_1$ and $\sigma_2$. Verify $\det( J(\phi))\gt 0$ and find $T_p(S)$. 

 A: $\sigma_1$ and $\sigma_2$ do not form an atlas. (The lower right portion of the sphere has no coordinate chart, for instance). But they're part of an atlas. 
The first question is "What's the domain for the transition function? In other words, for what pairs $(u, v)$ are $\bar{u}$ and $\bar{v}$ defined?  {I'm using bars because I don't know how to make tildes over the symbols). Well, it sure looks as if the second coordinate needs to be positive, because it's the result of taking a square root. So we know $v > 0$. (In your drawing, it looks as if you've got the positive-$y$ coordinate increasing to the LEFT, which is a bit weird, but I can work with that). So let's take a point $(u, v)$ with $u^2 + v^2 < 1$ and $v > 0$, and apply $\sigma_1$ to it. We get
$ \sigma_1(u, v) = (u, v, \sqrt{1 - u^2 - v^2} )$.
Now we want to apply $\sigma_2^{-1}$ to that, i.e., we want to find $(\bar{u}, \bar{v})$ such that $\sigma_2(\bar{u}, \bar{v})$ is the same as the displayed expression above. Well, 
$\sigma_2(\bar{u}, \bar{v}) = (\bar{u}, \sqrt{1 - \bar{u}^2 - \bar{v}^2}, \bar{v})$
and for this to be equal to the previous expression requires that they be term-by term equal, i.e., that 
$\bar{u} = u$
$\sqrt{1 - \bar{u}^2 - \bar{v}^2} = v$, and
$ \bar{v} = \sqrt{1 - u^2 - v^2}$.
Working with just the first two, we get 
$\bar{u} = u$ and
$\sqrt{1 - \bar{u}^2 - \bar{v}^2} = \sqrt{1 - u^2 - \bar{v}^2} = v$ 
which we can solve to get 
$1 - u^2 - \bar{v}^2 = v^2$, or
$\bar{v}^2 = 1 - v^2 - u^2$, or
$\bar{v} = \sqrt{1 - v^2 - u^2}$.
Why did I take the positive square root in that last formula? Because on the domain where the two coordinate charts overlap, we also have $\bar{v} > 0$. 
OK: so if you know $u$ and $v$, then the formulas for $\bar{u}$ and $\bar{v}$ are just
$(\bar{u}, \bar{v}) = (u, \sqrt{1 - v^2 - u^2})$
and that's your transition function: 
$\phi(u, v) = (u, \sqrt{1 - v^2 - u^2})$.
(It's possible that your text defines the transition to go in the other direction, but with this example in front of you, you should be able to do that as well). 
Computing the $2 x 2$ derivative matrix for the function $\phi$ should be just an exercise in calculus for you. And showing the det of that matrix is positive...well, it'll be straightforward, I promise. 
To describe $T_P(S)$, we need to specify the point $P$. I'm going to start with some $(u, v)$-pair satisfying $u^2 + v^2 < 1$ (i.e., in the domain of $\sigma_1$, which I'll call $U$), and say that $P = \sigma_1(u, v)$. 
Next, I'm going to observe that for any smooth curve $\gamma : [-a, a] \rightarrow S$ 
with $\gamma(0) = P$, there's a number $0 < b \le a$ and a curve $\beta : [-b, b] \rightarrow U$ such that $\gamma(t) = \sigma_1(b(t))$ for $t \in (-b, b)$. Why? Let $Q = \sigma_1(U)$. That's an open set on $S$. The preimage $\gamma^{-1}(Q)$ is therefore an open set in the reals, and it contains $0$. It therefore contains some open interval $(c, d)$ around $0$. Let $b = \min(|c|, |d|)$ and you've got what you need: the image of $\gamma$ on the interval $(-b, b)$ lies entirely in the image of $\sigma_1$. $\sigma_1$ is  1-1 onto its image. So for any $t \in (-b, b)$, there point $\gamma(t)$ in the image of $\sigma_1$ corresponds to a unique point $(u, v)$ in $U$. Call this point $\beta(t)$. Whew!
(This, by the way, is why we often work with coordinate charts, which go from $S$ to ${\mathbf R}^2$, rather than parametrizations, like your $\sigma_1$ and $\sigma_2$. If we had a coordinate chart, I'd just say $\beta = chart \circ \gamma$.
Anyhow, the point is that every smooth curve on $S$ passing through $P$ corresponds to a smooth curve in $\mathbf{R}^2$ passing through the point $(u, v)$. Let's look at just two of those:
$\beta_1(t) = (u + t, v)$
$\beta_2(t) = (u, v+t)$
When we map these to $S$ by applying $\sigma_1$, we get
$\gamma_1(t) = \sigma_1(\beta_1(t)) = (u+t, v, \sqrt{1 - (u+t)^2 - v^2})$ and
$\gamma_2(t) = \sigma_1(\beta_2(t)) = (u, v+t, \sqrt{1 - u^2 - (v+t)^2})$.
Let's compute their derivatives:
$\gamma_1'(t) = (1, 0, \frac{-1}{\sqrt{1 - (u+t)^2 - v^2}} (u+t))$ and
$\gamma_2'(t) = (0, 1, \frac{-1}{\sqrt{1 - u^2 - (v+t)^2}} (v+t))$.
These derivatives, at $t = 0$, give two tangent vectors:
$\gamma_1'(0) = (1, 0, \frac{-u}{\sqrt{1 - (u+t)^2 - v^2}})$ and
$\gamma_2'(t) = (0, 1, \frac{-v}{\sqrt{1 - u^2 - (v+t)^2}})$.
Your tangent plane must contain both of these. So you need an equation of the form 
$ Ax + By + Cz = 0$
that they both satisfy. Here's one:
$A = \frac{-u}{\sqrt{1 - (u+t)^2 - v^2}}$
$B = \frac{-v}{\sqrt{1 - (u+t)^2 - v^2}}$
$C = 1$.
If you correctly computed the Jacobian, then these two tangent vectors should have been the first and second columns of your Jacobian matrix, and the coefficient vector $[A, B, C]$ for the plane turns out to be just a multiple of the cross-product of those two columns. This works in general for surfaces, but for higher dimensions you need a generalized cross-product, and besides, it only works in the case where your $n$-manifold is embedded or immersed in dimension $n+1$. 
