$\newcommand{\inprod}[2]{\left\langle{#1}\,|\,{#2}\right\rangle}$ $\newcommand{\eqd}{\triangleq}$ $\newcommand{\setn}[1]{{\left\{{#1}\right\}}}$ $\newcommand{\setu}{\cup}$ $\newcommand{\brp}[1]{{\left(#1\right)}}$ $\newcommand{\brbl}[1]{\left\{#1\right.}$ How do they come about the difference equation when demonstrating a Strömberg wavelet $S(t)$ is orthogonal to a simple tent function $$\lambda_{\sigma}(t) \eqd \brbl{\begin{array}{ll} 1 & \text{if $t=\sigma$} \\0 & \text{if $t\in A_0$ and $t\neq\sigma$} \\\text{linear} & \text{otherwise} \end{array}}$$ in the computation of the Strömberg Wavelet? Here, $$\begin{array}{lcll} A_0 &\eqd& \setn{\cdots,-\frac{3}{2},-1,-\frac{1}{2},0,1,2,3,\cdots}&\text{and moreover}\\ A_1 &\eqd& A_0 \setu \setn{\frac{1}{2}} \end{array}$$ In particular, how can we arrive at this set of equations? $$0 = \inprod{S}{\lambda_{\sigma }} = \brbl{\begin{array}{l ll} S\brp{\sigma -1}+ 4S\brp{\sigma}+S\brp{\sigma +1} & \text{for $\sigma\in\setn{2,3,4,\ldots}$} & \text{(1.10)} \\ S\brp{\sigma-\frac{1}{2}}+ 4S\brp{\sigma}+S\brp{\sigma+\frac{1}{2}} & \text{for $\sigma\in\setn{-\frac{1}{2},-1,-\frac{3}{2},-2,\ldots}$} & \text{(1.11)} \\ 2S\brp{-\frac{1}{2}} + 9S\brp{0} + 6S\brp{\frac{1}{2}} + S\brp{1} & \text{for $\sigma=0$} & \text{(1.12)} \\ S\brp{0} + 6S\brp{\frac{1}{2}} + 13S\brp{1} + 4S\brp{2} & \text{for $\sigma=1$} & \text{(1.13)} \end{array}}$$
You can check the equations (1.10)–(1.13) here.
$\newcommand{\inprod}[2]{\left\langle{#1}\,|\,{#2}\right\rangle}$ $\newcommand{\eqd}{\triangleq}$ $\newcommand{\setn}[1]{{\left\{{#1}\right\}}}$ $\newcommand{\setu}{\cup}$ $\newcommand{\brp}[1]{{\left(#1\right)}}$ $\newcommand{\brs}[1]{{\left[#1\right]}}$ $\newcommand{\brbl}[1]{\left\{#1\right.}$ $\newcommand{\intcc} [2] {{\left[#1:#2\right]}}$ $\newcommand{\dx}{{\;dx}}$ I don't have a huge background in Strömberg Wavelets$\ldots$ but assuming the equation for (1.10) is correct, it should be possible to prove it by writing down the equations for $\Lambda_\sigma$ and $S(\sigma)$ and integrate their product over $\sigma-1\leq t\leq\sigma+1$. In particular, note that for $\sigma\in\setn{2,3,4,\ldots}$ $$\Lambda_{\sigma}(x) = \brbl{\begin{array}{ll} 1-\sigma+x & \text{for $\sigma-1\leq x\leq\sigma$} \\1+\sigma-x & \text{for $\sigma\leq x\leq\sigma+1$} \\0 & \text{otherwise} \end{array}}$$ and $$S(x) = \brbl{\begin{array}{ll} \frac{S(\sigma)-S(\sigma-1)}{1}\brp{x-\sigma}+S(\sigma) & \text{for $\sigma-1\leq x\leq\sigma$} \\\frac{S(\sigma+1)-S(\sigma)}{1}\brp{x-\sigma}+S(\sigma) & \text{for $\sigma\leq x\leq\sigma+1$} \end{array}}$$ Then $$\begin{align*} \inprod{S(x)}{\lambda_\sigma(x)} &= \int_{x\in\mathbb{R}}S(x)\lambda_\sigma(x)dx && \text{by definition of $\inprod{f(x)}{g(x)}$} \\&= \int_{x=\sigma-1}^{x=\sigma+1}S(x)\lambda_\sigma(x)dx && \text{because support of $\lambda_\sigma$ is $\intcc{\sigma-1}{\sigma+1}$} \\&= \int_{x=\sigma-1}^{x=\sigma}S(x)\lambda_\sigma(x)dx + \int_{x=\sigma}^{x=\sigma+1}S(x)\lambda_\sigma(x)dx \end{align*}$$
Actually, for the case $\sigma\in\setn{2,3,4,\ldots}$, I get a similar result as the linked reference, but different by a factor of $\frac{1}{6}$:
(lemma 1): $$ \begin{align} &\int_{x=\sigma-1}^{x=\sigma} \brs{ (A-B)(x-\sigma)+A } (1-\sigma+x) \dx \\&= \int_{x=\sigma-1}^{x=\sigma} \brs{ A(1-\sigma+x) + B(\sigma-x) } (1-\sigma+x) \dx \\&= \int_{x=\sigma-1}^{x=\sigma} A\brs{ x^2 + 2x(1-\sigma) + (1-\sigma)^2 } + B\brs{-x^2 + x(\sigma+\sigma-1) + \sigma(1-\sigma)} \dx \\&= \brs{ A\brs{ \frac{1}{3}x^3 + x^2(1-\sigma ) + x(1-\sigma)^2 } + B\brs{-\frac{1}{3}x^3 + \frac{1}{2} x^2(2\sigma-1) + x\sigma(1-\sigma)}}_{x=\sigma-1}^{x=\sigma} \\&= A\brs{ \frac{1}{3}\brp{\sigma^3-\brp{\sigma^3-3\sigma^2+3\sigma-1}} + \brp{\sigma^2-\brp{\sigma^2-2\sigma+1}}(1-\sigma ) + (1-\sigma)^2\brp{\sigma-(\sigma-1)} } \\&+ B\brs{-\frac{1}{3}\brp{\sigma^3-\brp{\sigma^3-3\sigma^2+3\sigma-1}} + \frac{1}{2}\brp{\sigma^2-\brp{\sigma^2-2\sigma+1}}(2\sigma-1) + \sigma(1-\sigma)} \\&= A\brs{\brp{\sigma^2 - \sigma + \frac{1}{3} } +(2\sigma-1)(1-\sigma)+(-\sigma+1)^2} \\&+B\brs{-\brp{\sigma^2 - \sigma + \frac{1}{3}} + \frac{1}{2}(2\sigma-1)(2\sigma-1) + \sigma(-\sigma+1)} \\&=A\brs{\sigma^2( 1-2+1)+\sigma(-1+2+1-2)+\brp{ \frac{1}{3}-1+1}} \\&+B\brs{\sigma^2(-1+2-1) + \sigma( 1-1-1+1) + \brp{-\frac{1}{3}+\frac{1}{2}}} \\&= \boxed{\frac{1}{3}A + \frac{1}{6}B} \end{align}$$
(lemma 2): $$\begin{align} &\int_{x=\sigma}^{x=\sigma+1} \brs{ (C-A)(x-\sigma)+C } (1+\sigma-x) \dx \\&= \int_{x=\sigma}^{x=\sigma+1} \brs{ C(1-\sigma+x) + A(\sigma-x) } (1+\sigma-x) \dx \\&= \int_{x=\sigma}^{x=\sigma+1} C\brs{-x^2 + x(\sigma+\sigma+1) + \sigma(-\sigma-1) } \\&+ A\brs{ x^2 + x(-\sigma-1-1-\sigma) + (1+\sigma)^2} \dx \\&= \brs{ C\brs{-\frac{1}{3}x^3 + \frac{1}{2} x^2(2\sigma+1) - x\sigma(\sigma+1)} +A\brs{ \frac{1}{3}x^3 - x^2( \sigma+1) + x(\sigma+1)^2 }}_{x=\sigma}^{x=\sigma+1} \\&= C\brs{-\frac{1}{3}\brp{\brp{\sigma^3+3\sigma^2+3\sigma+1}-\sigma^3} + \frac{1}{2}\brp{\brp{\sigma^2+2\sigma+1}-\sigma^2}(2\sigma+1) - \sigma(\sigma+1) } \\&+ A\brs{ \frac{1}{3}\brp{\brp{\sigma^3+3\sigma^2+3\sigma+1}-\sigma^3} - \brp{\brp{\sigma^2+2\sigma+1}-\sigma^2}( \sigma+1) + (\sigma+1)^2 } \\&= C\brs{-\sigma^2 + \sigma - \frac{1}{3} + \frac{1}{2}(2\sigma+1)(2\sigma+1) - \sigma^2 -\sigma } \\&+ A\brs{ \sigma^2 + \sigma + \frac{1}{3} - (2\sigma+1)( \sigma+1) + \sigma^2 + 2\sigma + 1 } \\&= C\brs{\sigma^2(-1+2-1) + \sigma(-1+1+1-1) + \brp{-\frac{1}{3}+\frac{1}{2}}} \\&+ A\brs{\sigma^2( 1-2+1) + \sigma( 1-2-1+2) + \brp{ \frac{1}{3}-1+1}} \\&= \boxed{\frac{1}{6}C + \frac{1}{3}A} \end{align}$$
(definitions): Let $A\eqd S(\sigma)$, $B\eqd S(\sigma-1)$, and $C\eqd S(\sigma+1)$.
Then$\ldots$ $$\begin{align} \inprod{S(x)}{\Lambda_\sigma(x)} &= \int_{x\in\mathbb{R}}S(x)\Lambda_\sigma(x)\dx && \text{by definition of $\inprod{f(x)}{g(x)}$} \\&= \int_{x=\sigma-1}^{x=\sigma+1}S(x)\Lambda_\sigma(x)\dx && \text{because support of $\Lambda_\sigma$ is $\intcc{\sigma-1}{\sigma+1}$} \\&= \int_{x=\sigma-1}^{x=\sigma}S(x)\Lambda_\sigma(x)\dx + \int_{x=\sigma}^{x=\sigma+1}S(x)\Lambda_\sigma(x)\dx \\&\eqd \int_{x=\sigma-1}^{x=\sigma } \brs{ (A-B)(x-\sigma)+A } (1-\sigma+x) \dx \\&+ \int_{x=\sigma }^{x=\sigma+1} \brs{ (C-A)(x-\sigma)+C } (1+\sigma-x) \dx && \text{by (definitions)} \\&= \brs{\frac{1}{3}A + \frac{1}{6}B} + \brs{\frac{1}{6}C + \frac{1}{3}A} && \text{by (lemma 1) and (lemma 2)} \\&= \frac{1}{6}\brs{B + 4A + C} \\&\eqd \boxed{\frac{1}{6}\brs{S(\sigma-1) + 4S(\sigma) + S(\sigma+1)}} && \text{by (definitions)} \end{align}$$