Given a group $G$ an a subgroup $H<G$, a representation of $H$ on $V$ is a pair $(\rho, V)$ where $\rho \colon H \to \mathrm{GL}(V)$ where $V$ is a vector space over a field $K$. We can the construct an induced representation of $G$ by $$\mathrm{Ind}_H^G \, \rho = K[G] \otimes_{K[H]} V \, .$$
Is there a way to generalize this for an arbitrary group homomorphism $f \colon H \to G$ such that when applying the construction for subgroups and the inclusion map $i \colon H \to G$ one recovers the original induced representation? That is, if instead of $H$ being a subgroup of $G$ we consider only a group homomorphism between $H$ and G.
At first I thought that, at least for the algebraic construction for finite groups, one could just change $H$ for $\mathrm{im}\, f$ and get done with it, but seems a too naive approach to work in general.
Extensions of scalars works for general ring homomorphisms, so the induced representations must work for general group homomorphisms.
In order for the tensor product $K[G]\otimes_{K[H]} V$ to define a $K[G]$-module, all you need is that $K[G]$ is a $(K[G],K[H])$-bimodule. For this, you can use any group homomorphism $f:H\to G$ to define an action of $H$ on $G$ on the right by $g\cdot h=gf(h)$, and this will be compatible with the translation action of $G$ on itself on the left. Then it suffices to extend this bilinearly to get a bimodule, and then the tensor product $K[G]\otimes_{K[H]} V$ (where $g\otimes \rho(h)v=gf(h)\otimes v$) is a left $K[G]$-module, pretty much as in the case where $f$ is the inclusion of a subgroup.
Moreover, we still have a natural correspondance between maps $\mathrm{Ind}_H^G (V,\rho)\to (W,\sigma)$ and maps $(V,\rho) \to \mathrm{Res}_H^G(W,\sigma)=(W,\sigma \circ f)$, for all representations $(W,\sigma) $ of $G$.
You mention that you tried to take the same construction as in the case where $H$ is a subgroup but with $H$ replaced by $im(f)$. That is indeed a bit too simple to work, but it's not that far from correct either : if you factorize $f$ as a $H\to \frac{H}{\ker(f)}\to G$, then since $im(f)\cong \frac{H}{\ker(f)}$ your suggestion actually does half of the job! So it is enough to find the representation induced by the quotient map $H\to \frac{H}{\ker(f)}$, i.e. the $K\left[\frac{H}{\ker(f)}\right]$-module $K\left[\frac{H}{\ker(f)}\right]\otimes_{K[H]}V$. This is actually isomorphic to the module of coinvariants $V_{\ker(f)}$, which is defined as the quotient of $V$ by the ideal generated by terms of the form $v-\rho(k)v$, where $v\in V$ and $k\in \ker(f)$.
Thus you can construct $K[G]\otimes_{K[H]}V$ as $$\bigoplus_{g\operatorname{im}(f)\in G/\operatorname{im}(f)} V_{\ker(f)},$$ with the action defined as in the case where $H$ is just a subgroup.