This community wiki solution is intended to clear the question from the unanswered queue.
As Arnaud D. remarked in his comment, the definition of a horn in the book by Laures and Szymik is false. In fact, the $r$-th horn is the simplicial subset of $\Delta^n$ which has as $m$-simplices exactly the order preserving maps $f:[m] \to [n]$ having $r$ in the image.
Geometrically the $r$-th horn is obtained from $\Delta^n$ by removing two simplices, namely the only $n$-simplex $id : [n] \to [n]$ and the unique $(n-1)$-simplex $s_r : [n-1] \to [n]$ with $r \notin s_r([n-1])$ (the face opposite to $r$). For obviuus reasons some German authors also use the word "Trichter" instead of "Horn".
So having for example $f:\Lambda^3_2 \to X$ is having the data of
- 4 0-simplicies $x_0,x_1,x_2,x_3$,
- 6 1-simplicies $f_{0,1},f_{1,2},f_{2,3},f_{0,2},f_{0,3},f_{1,3}$ that we can interpret as 'morphisms' $f_{i,j}: x_i \to x_j$ for $i<j$ in $\{0<1<2<3\}$, meaning the two 0-faces of $f_{i,j}$ are $x_i$ and $x_j$,
- 3 2-simplicies $f_{0,2,3}, f_{0,1,2}, f_{1,2,3}$, ($f_{0,1,3}$ beeing the one missing), that we can interpret as homotopies witnessing compositions : for example $f_{0,2,3} : f_{2,3}\circ f_{0,2} \sim f_{0,3}$,
- no 3-simplex.
So the horn filling condition states that given that data you can find the missing 2-simplex $f_{0,1,3}$, and the missing 3-simplex $f_{0,1,2,3}$, this 3-simplex, i.e. a homotopy of compositions of homotopies : $$f_{0,1,2,3} : f_{0,2,3} \circ f_{0,1,2} \sim f_{0,1,3} \circ f_{1,2,3}$$
The first composition of homotopies, $f_{0,2,3} \circ f_{0,1,2}$, says that we can decompose $f_{0,3}$ as $f_{2,3} \circ f_{0,2}$ first and then decomposite further to get $f_{2,3} \circ (f_{1,2} \circ f_{0,1})$.
The second composition of homotopies, $f_{0,1,3} \circ f_{1,2,3}$ says we can decompose $f_{0,3}$ as $f_{1,3} \circ f_{0,1}$ first and then as $(f_{2,3} \circ f_{1,3}) \circ f_{0,1}$.
So this 3-simplex encodes assosciativiy of the compositions of 3 morphisms up to homotopy, a homotopy between $f_{2,3} \circ (f_{1,2} \circ f_{0,1})$ and $(f_{2,3} \circ f_{1,3}) \circ f_{0,1}$.
I like to see 1-simplicies $f_{i,i+1}$ as the morphisms I want to compose, and $f_{i,j}$ with $|i-j| > 1$ as a candidate for compositions, so here for example let's say $f_{0,1} = f$, $f_{1,2} = g$, $f_{1,3} = h$, then $f_{0,2}$ will be a candidate for the composition $g \circ f$, $f_{1,3}$ will be a candidate of the composition $h \circ g$, and $f_{0,3}$ a candidate for composition $h\circ g \circ f$, but this $f_{0,3}$ can a candidate for $h \circ g \circ f$ in two ways : $f_{1,3} \circ f_{0,1}$ i.e. $(h\circ g) \circ f$ or $f_{2,3} \circ f_{0,2}$ i.e. $h\circ (g \circ f)$. 2 simplices when they exist are ways to compose and 3 simplicies are ways to compose those compositions and so on.
4-horns can be interpreted in a similar way but it is tedious to write down and you need to have good vision in 4-dimensions.
Best Answer
Via the correspondence mentioned in the answer, the inclusion $\Lambda^n_k\hookrightarrow\Delta^n$ is the $n$-tuple of $n-1$-simplices $$ (d_1(\iota),\dots,\hat d_k(\iota),\dots,d_n(\iota)), $$ where $\iota:\Delta^n_{n-1}\to\Delta^n_{n-1}$ is the identity.
The composite $\Lambda^n_k\hookrightarrow \Delta^n\xrightarrow{y} Y$ iteslf is an $n$-tuple of $n-1$-simplices in $Y$: $$ (y_1,\dots,\hat y_k,\dots,y_n), $$ From the discussion above this tuple is explicitly given by $$ (d_1(y),\dots,\hat d_k(y),\dots,d_n(y)), $$
The result $d_i y = y_i$ follows. Please correct me for any errors or imprecisions.