Exact sequences of vector bundles/spaces and isomorphisms between exterior powers $$\DeclareMathOperator{\rk}{rk}$$

Remark. Given a short exact sequence of holomorphic vector bundles:
$$0\to E\to F\to G\to0,$$
we cannot, in general, write $F\cong E\oplus G$, but it holds that:
$$\det F\cong\det E\otimes\det G.$$
Indeed, at every point $x$, one has an isomorphism:
$$\Lambda^kE^\ast\otimes\Lambda^hF\to\Lambda^{h-k}G,$$
induced by the inner product.

This is an excerpt of my professor's Complex Geometry notes. Naturally, the exact sequence gives us:
$$\rk F=\rk E+\rk G.$$
Let's see if the ranks match in those isomorphisms. Let us start with the $\det$ part:
\begin{align*}
\rk\det F={}&\rk\Lambda^{\rk F}F=\binom{\rk F}{\rk F}=1=1\cdot1=\binom{\rk E}{\rk E}\cdot\binom{\rk G}{\rk G}={} \\
{}={}&\rk\Lambda^{\rk E}E\cdot\rk\Lambda^{\rk G}G=\rk\det E\cdot\rk\det G=\rk(\det E\otimes\det G),
\end{align*}
where the binom part is taken from Wikipedia, and is a consequence of the fact that a basis is given by wedges of the elements of a dual basis, and there are that many linearly independent such wedges, and the rank of a tensor product is deduced from the dimension of a tensor product of vector spaces which I can prove to be the product of the dimensions of the "factors". So things are OK here.
However, for the other part, we have a problem. Noting that $E^\ast$ has the same rank as $E$, we get:
\begin{align*}
\rk(\Lambda^kE^\ast\otimes\Lambda^hF)={}&\rk\Lambda^kE^\ast\cdot\rk\Lambda^hF=\binom{\rk E}{k}\cdot\binom{\rk F}{h}={} \\
{}={}&\binom{\rk E}{k}\cdot\binom{\rk G+\rk E}{h}, \\
\rk\Lambda^{h-k}G={}&\binom{\rk G}{h-k},
\end{align*}
and I cannot seem to find a way to prove this. Indeed, thinking combinatorially, this makes no sense, because the first number is the number of ways of choosing $k$ objects out of $\rk E$ objects multiplied by the number of ways of choosing $h$ objects out of $\rk E+\rk G$ ones, why should that equal the number of ways of choosing $h-k$ objects out of $\rk G$, which is the second number?
That isomorphism would yield the $\det$ part. Indeed, if I tensor by $\Lambda^kE$, I should get that the $\Lambda^kE^\ast$ cancels out, and so:
$$\Lambda^hF\cong\Lambda^kE\otimes\Lambda^{h-k}G,$$
which is the $\det$ part once $h=\rk F,k=\rk E$.
I switched to fibers, so I had a exact sequence:
$$0\to V\xrightarrow{\phi}W\xrightarrow{\psi}X\to0,$$
where $\phi$ is injective, $\psi$ is surjective and $\ker\psi=\operatorname{Im}\phi$, and tried to construct the isomorphism by giving a map on simple tensors. I chose a basis $e_1,\dotsc,e_m$ for $E$, then completed the set of $f_i:=\phi(e_i)$ (which is linearly independent due to injectivity of $\phi$) to a basis $f_1,\dotsc,f_m,f_{m+1},\dotsc,f_n$ of $W$, and then $g_i:=\psi(f_{m+i})$ are generators of $X$ by surjectivity of $\psi$ (images of basis $\{f_i\}$ minus the zeros from the basis of $\operatorname{Im}\phi=\ker\psi$) and so they are a basis because they are the right number. A simple tensor in $\Lambda^kV^\ast\otimes\Lambda^hW$ is $v\otimes w$ with $v\in\Lambda^kV^\ast,w\in\Lambda^hW$. Suppose $v,w$ are simple wedges. So:
$$v=\bigwedge_{\ell=1}^k\left(\sum a_{i_\ell j}e_j^{\ast\ast}\right),\qquad w=\bigwedge_{p=1}^h\left(\sum b_{i_pj}f_j^\ast\right).$$
I reduced myself to no double stars by the canonical injection. I thought of mapping:
$$v\otimes w\mapsto\prod_{q=1}^k\left[\left(\sum b_{i_qj}f_j^\ast\right)\circ\phi\left(\sum a_{i_qj}e_j\right)\right]\cdot\bigwedge_{p=k+1}^h\left(\sum b_{i_pj}f_j^\ast\right),$$
and extend by linearity. So I match the first $k$ "factors" of the simple wedge $w$ to the $k$ of $v$, and multiply the applications of a "factor" of $w$ to the $\phi$-image of the matching "factor" of $v$, and then I keep the remaining "factors" of $w$. Let me give an example in low dimensions. Let $h=2,k=1$. Then we have:
$$\left(\sum a_je_j\right)\otimes\left(\left(\sum b_{1j}f_j^\ast\right)\wedge\left(\sum b_{2j}f_j^\ast\right)\right)\mapsto\left(\sum b_{1j}f_j^\ast\right)\circ\phi\left(\sum a_je_j\right)\cdot\sum b_{2j}f_j^\ast.$$
The "factors" of $v$ here are simply $\sum a_je_j$ (a single "factor"), and those of $w$ are the two sums $\sum b_{1j}f_j^\ast,\sum a_{2j}f_j^\ast$, and the mapping takes the first sum, applies it to the $\phi$-image of $\sum a_je_j^\ast$, and multiplies the result by the second sum. In other words, $v\otimes(w_1\wedge w_2)\mapsto w_1(\phi(v))\cdot w_2$. This brings me to $\Lambda^{h-k}W$, though. I want to go to $\Lambda^{h-k}X$. So I thought of composing this with the map that, on simple wedges, will send each "factor" to its image via the following map, which I define on the dual basis and can be extended by linearity:
$$f_i^\ast\mapsto\begin{cases} (\psi(f_i)^\ast & \psi(f_i)\neq0 \\ 0 & \psi(f_i)=0 \end{cases}.$$
The complicated mapping thus produced seems to be surjective, since I should be able to get all simple wedges this way, and the map is obviously linear and well-deefined. Indeed, if I take $v_1\wedge\dotso\wedge v_{h-k}$, I find preimages $\tilde v_i$ via $\psi$ of the factors, and then $(e_1\wedge\dotso\wedge e_k)\otimes(f_1^\ast\wedge \dotso\wedge f_k^\ast\wedge\tilde v_1\wedge\dotso\wedge \tilde v_{h-k})$ should map to the element I wanted to find a preimage of. So if the dimensions matched, I would have found the isomorphism.
Another remark that suggests dimensions might not match is that this map is constructed via a map that can send stuff to zero: $f_1^\ast$ goes to zero via that map defined for the second step of the construction of the candidate isomorphism. f that map is not injective, how can a composition of it with something else suddenly become injective?
So I am asking:

Are those isomorphisms correct? How can I construct them? Is the above attempt the correct way or am I right that it is not injective? If so, what is the correct isomorphism that I should try to find? Or how do I prove those dimensions match, and find that isomorphism?

 A: Following @QiaochuYuan's comment, I asked my professor via mail if $k,h$ were meant as "for all" or as the ranks, and he replied that the map is defined in general, but is an isomorphism only if $\DeclareMathOperator{\rk}{rk}h=\rk F,k=\rk E$, and added that to his notes. So the question becomes:

Is the above map an isomorphism in such a case?

Let us see how the construction particularizes to this case. We have $\dim V=k,\dim W=k,\dim X=k-h$. $\Lambda^kV$ is generated by $v:=e_1\wedge\dotso\wedge e_k$, so a generic element of said space is $av$. $\Lambda^kW^\ast=\{aw:w=f_1^\ast\wedge\dotso\wedge f_h^\ast,a\in\mathbb C\}$, and $\Lambda^{k-h}X=\{ax:x=(\psi(f_{l+1}))^\ast\wedge\dotso\wedge(\psi(f_h))^\ast,a\in\mathbb C\}$. So we would have:
$$av\otimes bw\mapsto abf_1^\ast(\psi(e_1))\cdots f_k^\ast(\phi(e_k))\bigwedge_{i=k+1}^h(\psi(f_i))^\ast.$$
For $a=b=1$, given that the huge product of the $f_j^\ast(\phi(e_j))$ is actually one, we find $x$. This means the basis $v\otimes w$ is mapped to the basis $x$, making this an isomorphism.
Since he says «the interior product induces», I guess this is what he had in mind. So this answers my question.
Yeah, well, technically I proved a linear algebra statement, whereas I would need a holomorphic vector bundle isomorphism, but I think I will safely assume that doing this on each fiber gives said isomorphism, i.e. that these isomorphisms glue together into a biholomorphism. Oh, and show that there exist local basis sections that allow gluing beyond a single trivialisation, where I would probably assume to have common trivialisations for the three line bundles $E,F,G$.
