The theorem you have indeed is useful, provided you know what the tangent space to $N$ looks like/ if you can easily figure it out, and provided you know what $F,M$ should be. In your particular case, it should be natural to choose $M= \Bbb{R}^3$, equipped with the maximal atlas containing the identity chart $(\Bbb{R}^3, id)$. Then, of course, we take $N = S^2$, and lastly, it should be reasonable to define $F: \Bbb{R}^3 \to \Bbb{R}$ by
\begin{align}
F(x,y,z) = z^2
\end{align}
Then, clearly, $f = F|_N$.
Now, let's side-track slightly and see how to think of tangent vectors as actual elements of your model space, so that we can see how to apply this to your question. Let $X$ be a smooth manifold (I avoid $M$ because I already used it above) of dimension $n$. Let $p \in X$, and let $(U, \alpha)$ be any chart of $X$ containing the point $p$ (i.e $p \in U$). Then, one can construct an isomorphism of $T_p(X)$ onto $\Bbb{R}^n$ as follows: define the map $\Phi_{\alpha, p}: T_pX \to \Bbb{R}^n$ by
\begin{align}
\Phi_{\alpha, p}\left( [c] \right) &:= (\alpha \circ c)'(0)
\end{align}
By definition of the equivalence relation, this map is well defined (in fact it is pretty much using this map that you use to get a vector space structure on $T_pX$).
Now, if $g: X \to Y$ is a smooth map between smooth manifolds, and $(U, \alpha)$ is a chart about a point $p \in X$, and $(V, \beta)$ is a chart for $Y$ about the point $g(p)$, then we have the following commutative diagram:
$\require{AMScd}$
\begin{CD}
T_pX @>{g_{*p}}>> T_{g(p)}Y \\
@V{\Phi_{\alpha,p}}VV @VV{\Phi_{\beta, g(p)}}V \\
\Bbb{R}^n @>>{D(\beta \circ g \circ \alpha^{-1})_{\alpha(p)}}> \Bbb{R}^m
\end{CD}
Which I'll leave to you to verify that it is actually commutative. In words what this means is that to compute $g_{*,p}$, you can choose charts on the domain and the target, and instead consider the (familiar euclidean type) derivative of the chart-representative map $\beta \circ g \circ \alpha^{-1}$.
Now, a special case of interest is the following: we have $X$ as a submanifold of some $\Bbb{R}^l$, and we're given a point $p \in X$. How do we think of $T_pX$ in an intuitive way? Well, $p$ is an element of $X$ and hence $\Bbb{R}^l$, so it (trivially) lies in the identity chart $(\Bbb{R}^l, id)$. Thus, rather than thinking of the tangent space as $T_pX$, which set-theoretically consists of equivalence classes of curves (which is rather abstract and tough to compute with), consider instead the $\dim X$-dimensional subspace $\Phi_{id,p}\left( T_pX \right) \subset \Bbb{R}^l$. This is precisely the intuitive picture of tangent space you would have.
For example, if $X= S^2$ considered as a submanifold of $\Bbb{R}^3$, then for each $p \in X= S^2$, $\Phi_{id,p}(T_pS^2)$ will be the (translated) tangent plane $\{\xi \in \Bbb{R}^3: \langle \xi, p\rangle = 0 \}$ (which is precisely the usual intuitive picture you might have). To prove this in a rigorous manner, it will be much easier if you know that $S^2$ can be written as a level set, say $h^{-1}(\{1\})$, where $h: \Bbb{R}^3 \to \Bbb{R}$ is defined by $h(x,y,z) = x^2 + y^2 + z^2$, and that $T_pS^2 = \ker h_{*p}$, so that (by the above commutative diagram, and basic linear algebra),
\begin{align}
\Phi_{\text{id}_{\Bbb{R}^3}, p}(T_pS^2) =
\ker D(\text{id}_{\Bbb{R}} \circ h\circ \text{id}_{\Bbb{R}^3}^{-1})_{\text{id}_{\Bbb{R}^3}(p)} =
\ker(Dh_p)
\end{align} where $Dh_p : \Bbb{R}^3 \to \Bbb{R}$ is the usual derivative.
So, now back to your question. You seek all $p \in S^2$ such that $f_{*,p} = 0$, or equivalently, by your theorem, those $p \in S^2$ such that $T_pS^2 \subset \ker F_{*,p}$. Or equivalently, those $p \in S^2$ such that
\begin{align}
\Phi_{\text{id}_{\Bbb{R}^3}, p}(T_pS^2) \subset \Phi_{\text{id}_{\Bbb{R}^3}, p} \left(
\ker F_{*,p} \right) = \ker DF_p
\end{align}
So, to answer your question, you just have to compute $DF_{p}$ for all $p \in S^2$ (in the usual calculus sense), compute the kernel of this map, and then see whether the translated tangent plane to the sphere lies inside the kernel.
I believe that this final computational part isn't difficult so I'll leave this all to you. I felt it is more important to see the logic behind what kind of computation needs to be performed.
Remark:
In your particular question, you've been aided by the fact that the theorem you stated gives a nice short proof (after a bit of practice, the reasoning I explained above in gory detail will become natural, so you'll be able to directly jump to my paragraph above... so this really is a short solution). However, suppose that you didn't know about that theorem. Then how would you go about finding the set of $p$ where $f_{*p} = 0$?
Well, the answer is very simple and straight forward (perhaps algebraically more tedious if you don't remember the charts). The sphere $S^2$ is a manifold, and as such, it has charts. The sphere is so nice that it can be covered by $2$-charts, (if you use stereographic projection).
Consider the stereographic projection from the north-pole: let $U_N = S^2 \setminus \{(0,0,1)\}$, and $\sigma_N : U_N \to \Bbb{R}^2$
\begin{align}
\sigma_N(x,y,z) = \left( \dfrac{x}{1-z}, \dfrac{y}{1-z} \right)
\end{align}
Its inverse is $\sigma_N^{-1}: \Bbb{R}^2 \to U_N$
\begin{align}
\sigma_N^{-1}(\xi,\eta) = \left( \dfrac{2\xi}{\xi^2 + \eta^2 +1}, \dfrac{2\eta}{\xi^2 + \eta^2 +1}, \dfrac{\xi^2 + \eta^2 - 1}{\xi^2 + \eta^2 +1} \right)
\end{align}
This chart covers the whole $S^2$ except the north-pole $(0,0,1)$. Now, it should be easy enough to verify that for $p \in U_N$, $f_{*,p} = 0$ if and only if $D(f \circ \sigma_N^{-1})_{\sigma_N(p)} = 0$. Or said differently, $f_{*, \sigma_N^{-1}(\xi,\eta)} = 0$ if and only if $D(f \circ \sigma_N^{-1})_{(\xi,\eta)} = 0$. But
\begin{align}
f \circ \sigma_N^{-1}(\xi, \eta) = \left( \dfrac{\xi^2 + \eta^2 - 1}{\xi^2 + \eta^2 +1} \right)^2
\end{align}
So, it should be easy to compute the standard derivative, and find where it vanishes. Then lastly, you just have to see if $f_{*, (0,0,1)} = 0$. To do this, you have to choose a chart which covers the north pole; you could use the stereographic projection from the south pole, or you could instead use the much simpler "graph chart" given by $V_{z,+} = \left\{(x,y,z) \in S^2| \, z>0 \right\}$ and $\psi_{z,+} : V_{z,+} \to \{(x,y)|\, x^2 + y^2 < 1\}$ given by
\begin{align}
\psi_{z,+}(x,y,z) = (x,y)
\end{align}
Note that I restricted the domain and target so that this is invertible, with inverse
\begin{align}
\psi_{z,+}^{-1}(x,y) = (x,y,\sqrt{1-x^2-y^2})
\end{align}
(I chose postive square root because of the definition of $V_{z,+}$). Hence, in this case,
\begin{align}
(f \circ \psi_{z,+}^{-1})(x,y) = 1-x^2-y^2
\end{align}
So, $f_{*, (0,0,1)} = 0$ if and only if $D(f \circ \psi_{z,+}^{-1})_{\psi_{z,+}(0,0,1)} = D(f \circ \psi_{z,+}^{-1})_{(0,0)} = 0$. Again, it should be easy enough to verify whether or not this condition is satisfied.
To recap: if you did not know that theorem, you just find an atlas for the manifold (and for computational purposes, find one with the fewest/simplest charts). Then, any property you wish to investigate about the push-forward $f_{*p}$ can be phrased equivalently in terms of the derivatives of the chart-representative maps, and solve the question in the chart (this is useful in general too).
For instance, if you hada slightly tougher question, say you're given some map $g: S^3 \to \Bbb{R}^4$, and you were asked to find all points where $g_{*p}$ had full rank, then I think a coordinate approach would be very mechanical and straight-forward. (As much as possible, it is a good idea to avoid charts, but it is also good to get used to them, because sometimes, they can provide a much quicker solution.)
Best Answer
Q1. Let $\gamma\colon \mathbb R\to M$ be a smooth curve and define $\gamma'(t_0) = d\gamma_{t_0}(\left.\frac d{dt}\right|_{t_0}).$ If $F\colon M\to N$ is a smooth map, then $$dF_{\gamma(t_0)}(\gamma'(t_0)) = dF_{\gamma(t_0)}(d\gamma_{t_0}(\left.\frac d{dt}\right|_{t_0})) = d(F\circ\gamma)_{t_0}(\left.\frac d{dt}\right|_{t_0}) = (F\circ\gamma)'(t_0) = \left.\frac d{dt}(F\circ\gamma)(t)\right|_{t=t_0}.$$
To get your expression, let $F = \mathrm{exp}\colon\mathfrak g\to G$ and $\gamma(t) = tX$. Notice that $\gamma'(0) = X$ (which you ask in Q2 why is true).
Q2. I'll use the same notation I've used above. Note that $\gamma'(0)\colon T_0\mathbb R\to T_0\mathfrak g\cong\mathfrak g.$ The key here is to understand that we identify vector space with its tangent space by assigning directional derivative to a vector. Thus, $$\gamma'(0)f = d\gamma_0(\left.\frac d{dt}\right|_{0})f = (\left.\frac d{dt}\right|_{0})(f\circ\gamma) = \left.\frac d{dt}f(tX)\right|_{t=0} \!\!\!\!= \lim_{t\to 0}\frac{f(0+tX)-f(0)}{t} = D_Xf(0) = Xf,$$ where the last equality is identification of $T_0\mathfrak g$ and $\mathfrak g$.
Q3. By the above, since $\gamma(0) = 0$ and $\gamma'(0) = X,$ we have
$$d(\mathrm{exp})_0(X) = d(\mathrm{exp})_0(\gamma'(0)) = \left.\frac d{dt}\mathrm{exp}(tX)\right|_{t = 0} = X.$$ Thus, $d(\mathrm{exp})_0$ is identity (if we identify $T_0\mathfrak g$ and $\mathfrak g$).