The Plucker relations are sufficient

As promised in the comment, here are some more details to Martin Brandenburg's proof. To get an understanding of what is going on, we first put ourselves in the situation that $s = \wedge^d\sigma$ for some surjective $\sigma\colon E\to F$. Why would we want to consider the map $t\colon \bigwedge\nolimits^{d+1}E\to E\otimes\bigwedge\nolimits^{d}F$ defined as \begin{align*}t(v_0\wedge\dots\wedge v_d) &= \sum_{k=0}^d(-1)^kv_k\otimes s(v_0\wedge\dots\wedge\widehat{v_k}\wedge\dots\wedge v_d) \\&=\sum_{k=0}^d(-1)^kv_k\otimes \sigma(v_0)\wedge\dots\wedge\widehat{\sigma(v_k)}\wedge\dots\wedge \sigma(v_d)\;\;? \end{align*} Well, I claim that its image is the kernel of the map $E\otimes\bigwedge^dF\xrightarrow{\sigma\otimes \mathrm{id}_{\wedge^d F}} F\otimes\bigwedge^dF$; since we can reconstruct $\sigma\colon E\to F$ up to isomorphism from its kernel, this map is very much relevant for what we are trying to do.

Let's prove the claim: It is easy to verify that $(\sigma\otimes \mathrm{id}_{\wedge^dF})\circ t$ factorises as $$\bigwedge\nolimits^{d+1}E\xrightarrow{\wedge^{d+1}\sigma}\bigwedge\nolimits^{d+1}F\to F\otimes\bigwedge\nolimits^{d}F,$$ where the latter map sends $v_0\wedge\dots\wedge v_d$ to $\sum_{k=0}^d(-1)^kv_k\otimes v_0\wedge\dots\wedge\widehat{v_k}\wedge\dots\wedge v_d$. But $\bigwedge\nolimits^{d+1}F = 0$; thus, $(\sigma\otimes\mathrm{id}_{\wedge^dF})\circ t = 0$, so that $\mathrm{im}(t)\subset \ker(\sigma\otimes \mathrm{id}_{\wedge^dF})$. Conversely, if $\sigma(w) = 0$, then $t(w\wedge v_1\wedge\dots\wedge v_d)=w\otimes s(v_1\wedge\dots\wedge v_d)$ and so $t$ maps surjectively onto $\ker(\sigma)\otimes\bigwedge^dF = \ker(\sigma\otimes \mathrm{id}_{\wedge^dF})$, as claimed.

What this means is that we have found a reasonable candidate for an inverse of the map $$\left\{E\xrightarrow{\sigma} F\to 0\right\}\to\left\{\bigwedge\nolimits^{d}E\xrightarrow{s}L\to 0\,\middle|\,\text{sat. Plücker}\right\},\sigma\mapsto \wedge^d\sigma,$$ by mapping $s$ to the cokernel of $T_s\colon \bigwedge\nolimits^{d+1}E\otimes L^\vee\xrightarrow{t\otimes \mathrm{id}_{L}}E\otimes L\otimes L^\vee\to E$, where the last map is just the natural isomorphism. The above shows that if we start with a $\sigma$, pass to $\wedge^d\sigma$, and then take the cokernel of $T_{\wedge^d\sigma}$, we get back $\sigma$ up to isomorphism. It remains to show that starting with some $s$ satisfying the Plücker relations, the candidate-inverse is well-defined (i.e., that the cokernel has rank $d$,) and that if we pass to the cokernel $\sigma\colon E\to F:=\mathrm{coker}(T_s)$ and then apply $\wedge^d$, we get back $s$ up to isomorphism. The latter is what those exact sequences are for, but we can phrase it without them:

We have two quotients of $\bigwedge^dE\otimes L$, namely, $\wedge^d\sigma\otimes \mathrm{id}_{L}\colon \bigwedge^dE\otimes L\to \bigwedge^dF\otimes L$ and $s\otimes \mathrm{id}_{L}\colon\bigwedge^dE\otimes L\to L\otimes L$ and we aim to show that they are isomorphic as quotients, i.e., that $\ker(\wedge^d\sigma\otimes\mathrm{id}_{L}) = \ker(s\otimes \mathrm{id}_{L})$. For this, we give nice presentations of those kernels.

For one, since $\ker(\sigma\otimes\mathrm{id}_{L})$ is the image of $t$, the kernel of $\wedge^d\sigma\otimes\mathrm{id}_{L}$ is the image of the map $\alpha\colon \bigwedge^{d-1}E\otimes\bigwedge^{d+1}E\to \bigwedge^{d}E\otimes L$, mapping $v\otimes w$ to $v\wedge t(w)$.

For the other map, note that $\ker(s\otimes \mathrm{id}_{L})=\ker(s)\otimes L$ is generated by elements of the form $v\otimes s(w)-w\otimes s(v)$, since, for $s(v)=0$ and $f = s(w)\in L$ arbitrary, $v\otimes s(w) - w\otimes s(v) = v\otimes f$. In particular, with $\beta\colon \bigwedge^dE\otimes\bigwedge^dE\to \bigwedge^dE\otimes L$ mapping $v\otimes w$ to $v\otimes s(w)- w\otimes s(v)$, we get $\ker(s\otimes \mathrm{id}_{L}) = \mathrm{im}{(\beta)}$.

Thus, if we manage to show that $(s\otimes \mathrm{id}_{L})\circ\alpha = 0$ and $(\wedge^d\sigma\otimes\mathrm{id}_{L})\circ\beta = 0$, then we conclude $$\ker(\wedge^d\sigma\otimes\mathrm{id}_{L})=\mathrm{im}{(\alpha)}\subset\ker(s\otimes \mathrm{id}_{L}) = \mathrm{im}{(\beta)}\subset \ker(\wedge^d\sigma\otimes\mathrm{id}_{L}),$$ which implies equality everywhere. In particular, $L\cong\bigwedge^dF$ as quotients of $\bigwedge^d E$ and so $\bigwedge^dF$ is invertible, hence $F$ has rank $d$; this is all we wanted to show.

Finally, we show the two identities $(s\otimes \mathrm{id}_{L})\circ\alpha = 0$ and $(\wedge^d\sigma\otimes\mathrm{id}_{L})\circ\beta = 0$. Tracing through the definitions shows that the former is the Plücker relation, and that the second is equivalent to $\wedge^d\sigma v\otimes s(w) = \wedge^d\sigma w\otimes s(v)$ for all $v,w\in\bigwedge^dE$. That is, we want $\wedge^d\sigma\otimes s$ to be symmetric. By construction of $\sigma$, we always have $$0 = \sum_{k=0}^d(-1)^{k}\sigma w_k\otimes s(w_0\wedge\dots\wedge \widehat{w_{k}}\wedge \dots\wedge w_d),$$ and so the symmetry of $\wedge^d\sigma\otimes s$ follows from what M. Brandenburg calls the Symmetry Lemma (4.4.15); I have nothing to add to his proof of this lemma.

I cannot put this as a comment and I'm sorry that it is not a complete answer, but a good introduction to Grassmanninans is written by Gathmann: http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2014/alggeom-2014-c8.pdf

In particular the answer to your second question could be Corollary 8.13 (look at the proof).