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})) $.
If we replace the basis $ \{x_{1},x_{2}\} $ with the basis $ u_{1}:=x_{1}+x_{2} $ and $ u_{2}:= x_{2}-x_{1} $,then the derivation $ \delta $ becomes $ u_{1} \frac{\partial}{\partial u_{1}}-u_{2}\frac{\partial}{\partial u_{2}} $. The co-action of $ \widehat{\mathbb{G}_{a}} $ on $ \operatorname{Spf}(k[[u_{1},u_{2}]]) $ sends $ u_{1} $ to $ e^{t} u_{1} $ and $ u_{2} $ to $ e^{-t} u_{2} $.
Suppose that there exists a $ g(x_{1},x_{2}) $$ g(u_{1},u_{2}) \in k[[u_{1},u_{2}]] $ such that $ \beta^{\sharp}(g(x_{1},x_{2})) = g(x_{1},x_{2})+th(x_{1},x_{2}) $$ \beta^{\sharp}(g(u_{1},u_{2})) = g(u_{1},u_{2})+h(u_{1},u_{2})t $ for some $ h(x_{1},x_{2}) \in k[[x_{1},x_{2}]]^{\widehat{\mathbb{G}_{a}}} $, then let$ h(u_{1},u_{2}) \in k[[u_{1},u_{2}]]^{\widehat{\mathbb{G}_{a}}} $.
Let $ g(x_{1},x_{2}) $$ g(u_{1},u_{2}) $ equal $ \sum_{j=0}^{\infty} g_{j}(x_{1},x_{2}) $$ \sum_{j=0}^{\infty} g_{j}(u_{1},u_{2}) $ where $ g_{j}(x_{1},x_{2}) $$ g_{j}(u_{1},u_{2}) $ is a homogeneous polynomial in $ k[u_{1},u_{2}] $ of degree $ j $ in $ k[x_{1},x_{2}] $. The ring Also, let $ k[[x_{1},x_{2},t]] $ has a bi-grading where \begin{align*} \deg(x_{1}) &= (1,0) \\ \deg(x_{2}) &= (1,0) \\ \deg(t) &= (0,1). \end{align*} If the reader takes for granted that the requirement that$ g_{j}(u_{1},u_{2}) $ equal $ \sum_{i=0}^{j} a_{i,j} u_{1}^{i} u_{2}^{j-i} $. Under the bico-degreeaction $ (J,\ell) $ part equal zero whenever$ \beta^{\sharp}(g_{j}(u_{1},u_{2})) = \sum_{i=0}^{j} a_{i,j} e^{(2i-j)t} u_{1}^{i}u_{2}^{j-i} $.
If $ \ell>1 $ adds infinitely many non-trivial$ \delta(g(u_{1},u_{2})) = \beta^{\sharp}(g(u_{1},u_{2}))-g(u_{1},u_{2}) $, linear conditions on the space of coefficientsthen \begin{align*} h(u_{1},u_{2})t &= \beta^{\sharp}(g(u_{1},u_{2}))-g(u_{1},u_{2}) \\ &= \sum_{j=0}^{\infty} \sum_{i=0}^{j} a_{i,j} (e^{(2i-j)t}-1)u_{1}^{i}u_{2}^{j-i}. \end{align*}
If $ [a_{0}:\cdots:a_{J}] \in \mathbb{P}^{J}_{k} $ such that$ h(u_{1},u_{2}) $ is equal to $ g_{J}(x_{1},x_{2}) $ equals$ \sum_{j=0}^{\infty} h_{j}(u_{1},u_{2}) $ where $ \sum_{c=0}^{J} a_{c}x_{1}^{c}x_{2}^{J-c} $$ h_{j}(u_{1},u_{2}) $ is a degree $ j $, homogeneous polynomial in $ k[u_{1},u_{2}] $, then one arrives at the conclusionfor any $ j $ such that the answer$ h_{j}(u_{1},u_{2}) $ is nonon-zero, \begin{align*} k[u_{1},u_{2}][t] & \ni h_{j}(u_{1},u_{2})t \\ &= \sum_{i=0}^{j} a_{i,j} e^{(2i-j)t} u_{1}^{i} u_{2}^{j-i} \\ & \notin k[u_{1},u_{2}][t]. \end{align*} As a result, such a $ g(u_{1},u_{2}) $ cannot exist for this rather simple action of $ \widehat{\mathbb{G}_{a}} $ on $ \operatorname{Spf}(k[[x_{1},x_{2}]]) $.
