A hypersubstitution $\sigma$ is (see, for example, Universal Algebra and Applications in Theoretical Computer Science, by Denecke and Wismath) mapping from term $f_i(x_1,...,x_{n_i})$ to the term $\sigma(f_i)$. It follows that every hypersubstitution of type $\tau$ induces a mapping $\hat{\sigma}:W_\tau(X)\to W_\tau(X)$ as follows:for any $w\in W_\tau(X)$, the term $\hat{\sigma}[w]$ is defined by
(1) $\hat{\sigma}[x]:=x$ for any variable $x\in X$
(2) $\hat{\sigma}[f_i(w_1,...,w_{n_i})]:=\sigma(f_i)(\hat{\sigma}[w_1],...,\hat{\sigma}[w_{n_i}]).$
From the book M-Solid Varieties of Algebras, by Koppitz and Denecke, we have
Definition 3.1.1 Let $W_{\tau}(X_m)$ and $W_{\tau}(X_n)$ be the sets of all $m$-ary and $n$-ary terms of type $\tau$, for $1 \leq m,n \in \mathbb N$. Then the operation $S_m^n:W_{\tau}(X_n)\times W_{\tau}(X_m)^n \to W_{\tau}(X_m)$ is defined inductively as follows:
(i) $S_m^n(x_i,t_1,\ldots,t_n)=t_i$, $\;x_i \in X_n$, $t_1, \ldots,t_n \in W_{\tau}(X_m)$,
(ii) $S_m^n(f_i(s_1, \ldots,s_{n_i}),t_1,\ldots,t_n)=f_i(S_m^n(s_1,t_1,\ldots,t_n),\ldots,S_m^n(s_{n_i},t_1,\ldots,t_n)),\;$ $f_i(s_1\ldots,s_{n_i})\in W_{\tau}(X_n)$.
And then
Definition 3.1.3 Let $\sigma:\{f_i:i \in I\} \to W_{\tau}(X)$ be a mapping assigning to every $n_i$-ary operation symbol $f_i$ of type $\tau$ an $n_i$-ary operation term $\sigma(f_i)$. Any such map $\sigma$ will be called an hypersubstitution of type $\tau$.
[...] Every hypersubstitution of type $\tau$ induces a mapping $\hat{\sigma}:W_{\tau}(X)\to W_{\tau}(X)$ on the set of all terms of type $\tau$ as follows. For any term $t \in W_{\tau}(X)$, the term $\hat{\sigma}[t]$ is defined inductively by
(1) $\hat{\sigma}[x]=x$ for any variable $x \in X$ and
(2) $\hat{\sigma}[f_i(t_1,\ldots,t_{n_i})]=S_n^{n_i}(\sigma(f_i),\hat{\sigma}[t_1],\ldots,\hat{\sigma}[t_n])$.
How to reconcile the two definitions of hypersubstitution (why are they equivalent)?
Notice that it all amounts to prove that $$S_n^{n_i}(\sigma(f_i),\hat{\sigma}[t_1],\ldots,\hat{\sigma}[t_{n_i}]) = \sigma(f_i)(\hat{\sigma}[t_1],\ldots,\hat{\sigma}[t_{n_i}]).$$ Let $x_1,\ldots,x_{n_i} \in X$. Then $$S_n^{n_i}(\sigma(f_i)(x_1,\ldots,x_{n_i}),\hat{\sigma}[t_1],\ldots,\hat{\sigma}[t_{n_i}]) =\sigma(f_i)(S_n^{n_i}(x_1,\hat{\sigma}[t_1],\ldots,\hat{\sigma}[t_{n_i}]),\ldots,S_n^{n_i}(x_{n_i},\hat{\sigma}[t_1],\ldots,\hat{\sigma}[t_{n_i}])),$$ by Definition 3.1.1 (ii), $$=\sigma(f_i)(\hat{\sigma}[t_1],\ldots,\hat{\sigma}[t_{n_i}]),$$ by Definition 3.1.1 (i).