Like often when dealing with sheaves/vector bundles, this is a two-step process. First define maps, then check their properties (exact sequence/isomorphisms...) The first step needs a global construction, for the second, we will be able to work locally.
Step 1 : this is hard to define properly a map $\bigwedge^p F\rightarrow \bigwedge^{p-1}E\otimes L$ since we don't have maps $F\rightarrow E$, so we will proceed differently. Let $C$ be the cokernel of $\bigwedge^pE\rightarrow\bigwedge^pF$. Tensor the exact sequence $0\rightarrow E\rightarrow F\rightarrow L\rightarrow 0$ by $\bigwedge^{p-1}E$ and construct the diagram :
$$
\require{AMScd}
\begin{CD}
0@>>>\bigwedge^{p-1}E\otimes E@>>>\bigwedge^{p-1}E\otimes F@>>>\bigwedge^{p-1}E\otimes L@>>>0\\
{}@VVV@VVV\\
0@>>>\bigwedge^pE@>>>\bigwedge^pF@>>>C@>>>0
\end{CD}
$$
where the first vertical map is the usual product in the exterior algebra $$(e_1\wedge...\wedge e_{p-1})\otimes e_p\mapsto e_1\wedge...\wedge e_{p-1}\wedge e_p $$ and the second is the composition of $\bigwedge^{p-1}E\otimes F\rightarrow\bigwedge^{p-1} F\otimes F\rightarrow F$, in other words : $$(e_1\wedge...\wedge e_{p-1})\otimes f_p\mapsto i(e_1)\wedge...\wedge i(e_{p-1})\wedge f_p$$
Clearly, the square is commutative, hence we have an induce map $\bigwedge^{p-1}E\otimes L\longrightarrow C$.
Step 2 : let us show that this map is an isomorphism. This is enough to work locally. In this case, we can assume that $F\simeq E\oplus L$, and that $E$ and $L$ are free with basis $e_1,...,e_n$ and $l$. We have induced basis for $F$ (written by abuse of notations $(e_1,...,e_n,l)$), basis for every exterior power of $E$ (for $\bigwedge^k E$, this is the family $(e_I)_I$ with $I\subset\{1,...,n\}$ is of cardinality $k$) and for every exterior power of $F$ (for $\bigwedge^k F$, this is the family $(e_J)_J\cup(e_I\wedge l)$ with $I,J\subset\{1,...,n\}$ is of cardinality $k-1$ and $k$ respectively).
Injectivity : assume $\sum a_I e_I\otimes l$ is in the kernel. By diagram chasing, $\sum a_I e_I\wedge l$ is an element of $\bigwedge^pE$, but this is not possible because $e_I\wedge l$ are linearly independent and are not $\bigwedge^p E$, unless all the $a_I$ are zero.
Surjectivity : let $c\in C$ and $\sum a_J e_J+\sum b_I e_I\wedge l$ a lift in $\bigwedge^p F$ (here the sum is over all $J\subset\{1,...,n\}$ of cardinality $p$ and all $I\subset\{1,...,n\}$ of cardinality $p-1$). Clearly, $\sum b_I e_I\wedge l$ is another lift of $c$. This shows that $c$ is the image of $\sum b_I e_I\otimes l$.
Hence the vertical map is an isomorphism and this concludes the proof.
You proposed another map for $\bigwedge^p F\rightarrow \bigwedge^{p-1}E\otimes L$. Let us see if this is indeed the same :
We only need to compute your map and mine on our previous basis.
Let $J=\{i_1,....,i_p\}\subset\{1,...,n\}$ with $i_1<...<i_p$, then $e_J=e_{i_1}\wedge...\wedge e_{i_p}$. With the previous construction, this is an element of $\bigwedge^p E$ so this element is zero in $C\simeq\bigwedge^{p-1}E\otimes L$. With your definition, this is also $0$.
Let $I=\{i_1,....,i_{p-1}\}\subset\{1,...,n\}$ with $i_1<...<i_{p-1}$. Then $e_I\wedge l=e_{i_1}\wedge...\wedge e_{i_{p-1}}\wedge l$. With the previous construction, this maps to $e_{i_1}\wedge...\wedge e_{i_{p-1}}\otimes l$. With your construction, this maps to
$$\sum_k (-1)^k e_{i_1}\wedge...\wedge\widehat{e_{i_k}}\wedge...\wedge e_{i_{p-1}}\wedge l\times 0+(-1)^p e_{i_1}\wedge...\wedge e_{i_{p-1}}\otimes l$$
Hence your map is the same as mine, up to the $(-1)^p$ sign.
Note : working instead with the sequence $0\rightarrow \bigwedge^p E\rightarrow\bigwedge^p F\rightarrow L\otimes \bigwedge^{p-1}E\rightarrow 0$ remove the sign.
Finally, some kind of interpretation : $F$ is made of $E$ and $L$. Thus $\bigwedge^p F$ is made of $\bigwedge^p E$ (clearly a subobject) and something else. If $F$ splits as $F\simeq E\oplus L$, then we would have decomposition $f_i=e_i+l_i$ so that $$f_1\wedge...\wedge f_p=(e_1+l_1)\wedge...\wedge (e_p+l_p)=(e_1\wedge...\wedge e_p)+\sum_k e_1\wedge...\wedge l_k\wedge ...\wedge e_p + 0$$
The zero comes from the fact that $L$ is of dimension $1$ so every product $l_i\wedge l_k$ above will be zero. Since you can change the order of terms (with a sign), you can put the $l_i$ at the end, so every element $f_1\wedge...\wedge f_p$ is a sum of an element of $\bigwedge^p E$ and an element of the form $e_1\wedge...\wedge e_{p-1}\wedge l$. It turns out (since we can produce adequate morphisms) that the part of $\bigwedge^p F$ made by the elements of the form $e_1\wedge...\wedge e_{p-1}\wedge l$ is a quotient which is $\bigwedge^{p-1}E\otimes L$.
More generally, if $L$ is not a line bundle anymore (write it $G$), the decomposition $\bigwedge^p(E\oplus G)=\bigoplus_{i+j=p}\bigwedge^i E\otimes\bigwedge^j G$ still has an interpretation if the sequence $0\rightarrow E\rightarrow F\rightarrow G\rightarrow 0$ does not split. Indeed, the bundle $\bigwedge^pF$ is now endowed with a filtration $F^k\bigwedge^pF$ such that $F^k\bigwedge^pF$ is the image of $E^{\otimes k}\otimes F^{p-k}\rightarrow F^p\rightarrow\bigwedge^p F$ (in other words, $F^k\bigwedge^pF$ is the subset of $\bigwedge^pF$ whose elements can be written with a product of $k$-elements from $E$). Now, the successive quotient are $F^k/F^{k+1}=\bigwedge^k E\otimes\bigwedge^{p-k} G$.
You are in the situation where the filtration is of only 2-steps :
$$ 0=F^{p+1}\bigwedge^pF\subset\bigwedge^p E=F^p\bigwedge^p F\subset F^{p-1}\bigwedge^p F=\bigwedge^p F$$
