Having $l$ for both leaves and left children might be confusing. Let $L,R$ be the number of left and right children in the binary tree.
There is indeed a bijection between ordered trees and binary trees,
called first child next sibling.
The nodes are kept the same, but the left binary child points to the first child
in the ordered tree, whereas the right child points to the next sibling.
(In both cases, insofar these nodes exist.)
The number of nodes in both tree equals $n$.
Any tree with $n$ nodes has $n-1$ children in total.
Thus the binary tree has a total of $n-1$ child-pointers also.
Each leaf in the ordered tree is a node without children.
Thus there are $\ell$ nodes without children, i.e., $\ell$ binary nodes without first child,
and the remaining $L=n-\ell$ nodes have a left pointer.
This accounts for $n-\ell$ children, out of a total $n-1$ children,
so the rest must be next siblings, $R=\ell-1$.
Is this, eeeuh, homework?