Given a finite group $G$, $H$ a subgroup, a field $K$, and a $K[H]$-module $M$, define
$$\operatorname{ind}_H^G M := \lbrace f : G \to M : f(gh) = h^{-1}f(g)\ \text{for all}\ g \in G, h \in H\rbrace.$$
This is a subspace of $\operatorname{Hom}(G, M)$. For all $g \in G$ define now $(gf)(g^\prime) = f(g^{-1}g^\prime)$, so that $gf \in \operatorname{ind}_H^G M$ also.
How does one now define a $K[G]$-module structure on $\operatorname{ind}_H^G M$?
Given the definition of $gf$ I'd assume
$$\left( \sum_{g \in G} a_g\varepsilon_g\right) \cdot f = \sum_{g \in G} a_g\cdot (\varepsilon_gf)$$
would be the appropriate action, just wanted to see if I can get some confirmation on this, since all other texts I've seen define the induced module directly as a tensor product.
You essentially have it correct in your comment. Given that $G$ acts on $\operatorname{Hom}(G,M)$ via $(g\cdot f)(g')=f(g^{-1}g')$, you want to define the $K[G]$-action as $(\epsilon_g\cdot f)(g')=f(g^{-1}g')$, since $\epsilon_g$ is essentially the "natural" copy of $g\in G$ in $K[G]$. Then extend $K$-linearly.
The minor issue with putting $(\epsilon_g\cdot f)(x)=f(gx)$ is that $$ (\epsilon_g\cdot(\epsilon_{g'}\cdot f))(x)=(\epsilon_{g'}\cdot f)(gx)=f(g'gx)=(\epsilon_{g'g}\cdot f)(x)=((\epsilon_{g'}\epsilon_g)\cdot f)(x) $$ but you need it to be equal to $((\epsilon_g\epsilon_{g'})\cdot f)(x)$. Acting on the left of the argument $x$ by the inverse gives the correct order of the factors acting on $f$.