When thinking about chain homotopies in a setting involving simplices it can be helpful to consider the product $\Delta^p\times I$ where $\Delta^p$ is a $p$-simplex and $I=[0,1]$. The formula $h\sigma=\sigma_{\sharp} b_p \ast(i_p-si_p-h\partial i_p)$ corresponds to a certain inductively defined subdivision of $\Delta^p\times I$ obtained by coning off a subdivision of $\Delta^p\times\partial I \cup \partial \Delta^p \times I$ to a point in the interior of $\Delta^p\times I$. The subdivision of $\Delta^p\times\partial I \cup \partial \Delta^p \times I$ is $\Delta^p$ itself (unsubdivided) on $\Delta^p\times \{0\}$ and the barycentric subdivision of $\Delta^p$ on $\Delta^p\times\{1\}$. These are the terms $i_p$ and $si_p$ in the formula. On $\partial \Delta^p\times I$ one uses the subdivision given by induction. This is the term $h\partial i_p$. The term $\sigma_{\sharp} b_p$ corresponds to the point in the interior of $\Delta^p\times I$ that one cones off to, with the symbol $\ast$ denoting the coning operation.
What is perhaps most puzzling is that the formula says nothing about taking the product with $I$, but this is because in reality one takes the subdivision of $\Delta^p\times I$ and projects it to $\Delta^p$ before applying the map $\sigma$, whose domain is $\Delta^p$ rather than $\Delta^p \times I$.
I have seen this method of subdividing $\Delta^p\times I$ in several books when they are developing homology theory, but it is more complicated than necessary. A simpler subdivision that suffices is to cone off a subdivision of $\Delta^p\times \{0\} \cup \partial \Delta^p \times I$ to the barycenter of $\Delta^p\times\{1\}$, where $\Delta^p\times \{0\}$ is unsubdivided and $\partial \Delta^p \times I$ has the subdivision given inductively. On $\Delta^p\times \{1\}$ this gives just the usual barycentric subdivision, which is also defined inductively. There is a picture of this subdivision of $\Delta^p\times I$ in the case $p=2$ on page 122 of my algebraic topology book. Perhaps other books such as the Lee book you mention don't give a picture because the picture would be more complicated for the more complicated subdivision. An advantage of the simpler subdivision is that the formula for $h\sigma$ becomes just $\sigma_{\sharp} b_p \ast(i_p-h\partial i_p)$, without the term $si_p$.
The more complicated formula is given in the classic book of Eilenberg and Steenrod (page 197) without pictures or explanation. Perhaps other books are just following suit.