Consider a representation $\;A:\,G\longrightarrow GL(\mathbb{V})\;$ of a group $G$ in a vector space $\mathbb{V}\,$: $$ A:\qquad G\times {\mathbb V} \longrightarrow {\mathbb V}\;,\quad (g,\,v)\longmapsto A_g\,v\;. $$ For a subgroup $K<G$, we get its representation on $\mathbb V$ by forgetting how to act with elements outside $K$: $$ A(G)\downarrow K:\qquad K\times {\mathbb V} \longrightarrow {\mathbb V}\;,\quad (k,\,v)\longmapsto A_k\,v\;. $$ Now, a special case. Suppose $\mathbb V$ is the space of all functions $\varphi$ mapping the elements of $G$ to another vector space $\mathbb W\,$: $$ {\mathbb{V}} = \left\{\varphi~\,{\Large{|}}\,~\varphi\,:~G\longrightarrow{\mathbb{W}}\right\}~, $$ while $A$ is acting by left translations: $$ \varphi(g^{\prime})\stackrel{g}{\longmapsto}\varphi(g^{-1}g^{\prime})\;, $$ where $g^{\prime}$ is a free variable.
In this situation, what exactly is implied by restriction $A(G)\downarrow K\,$?
Evidently, we are now acting only by elements $k\in K$. Should we restrict the domain accordingly? This is to say, will the restriction be acting as $$ \varphi(g^{\prime})\stackrel{k}{\longmapsto}\varphi(k^{-1}g^{\prime})\;,\quad k\in K $$ or as $$ \varphi(k^{\prime})\stackrel{k}{\longmapsto}\varphi(k^{-1}k^{\prime})\;,\quad k\in K\;, $$ $g^{\prime}\in G\,$ and $\,k^{\prime}\in K\,$ being free variables.
We may consider each $\varphi(k')$ defined on $K<G$ as a function which is defined on $G$ but is set to be zero on elements outside $K$. So such functions constitute a subspace $\,^{^K}{\mathbb V}\subset\mathbb V$. This enables me to reformulate my question: does the restriction $A(G)\downarrow K$ imply a restriction of the domain from $\mathbb V$ to $\,^{^K}{\mathbb V}$ ?
To turn my comment into an answer:
No, by your own definition, $\downarrow K$ does not change the vector space on which the group acts.
It is true that $^K \mathbb{V}$ is a subrepresentation of $A \downarrow K$.