The answer is no, such a ``local slice'' does not exist in general even for algebraic actions. Let $ \delta \in T_{k[x_{1},x_{2}]/k} $ be the derivation $ x_{1} \frac{\partial}{\partial x_{2}}+x_{2}\frac{\partial}{\partial x_{1}} $. By the infinite series identities \begin{align*} \cos(it)&=\sum_{s=0}^{\infty} t^{2s}/(2s)! \\ -i\sin(it) &= \sum_{s=0}^{\infty} t^{2s+1}/(2s+1)! \end{align*} the action of $ \widehat{\mathbb{G}_{a}} $ on $ \operatorname{Spf}(k[[x_{1},x_{2}]]) $ obtained from $ \delta $ is the action $ \beta $ which sends $ (t_{0},(a_{1},a_{2})) $ to $ (a_{1}\cos(it_{0})-ia_{2}\sin(it_{0}),a_{2}\cos(it_{0})-ia_{1}\sin(it_{0})) $.
Suppose there exists a $ g(x_{1},x_{2}) $ such that $ \beta^{\sharp}(g(x_{1},x_{2})) = g(x_{1},x_{2})+th(x_{1},x_{2}) $ for some $ h(x_{1},x_{2}) \in k[[x_{1},x_{2}]]^{\widehat{\mathbb{G}_{a}}} $, then let $ g(x_{1},x_{2}) $ equal $ \sum_{j=0}^{\infty} g_{j}(x_{1},x_{2}) $ where $ g_{j}(x_{1},x_{2}) $ is a homogeneous polynomial of degree $ j $ in $ k[x_{1},x_{2}] $. The ring $ k[[x_{1},x_{2},t]] $ has a bi-grading where \begin{align*} \deg(x_{1}) &= (1,0) \\ \deg(x_{2}) &= (1,0) \\ \deg(t) &= (0,1). \end{align*} If the reader takes for granted that the requirement that the bi-degree $ (J,\ell) $ part equal zero whenever $ \ell>1 $ adds infinitely many non-trivial, linear conditions on the space of coefficients $ [a_{0}:\cdots:a_{J}] \in \mathbb{P}^{J}_{k} $ such that $ g_{J}(x_{1},x_{2}) $ equals $ \sum_{c=0}^{J} a_{c}x_{1}^{c}x_{2}^{J-c} $, then one arrives at the conclusion that the answer is no for this rather simple action.