An extended comment: if $k$ has characteristic zero then there is no continuous embedding of local rings $k[[u,v]]\to k[[t]]$. Indeed, suppose it maps $(u,v)$ to $(U,V)$. Let $d$ be the valuation of $U$, that is, $u\in t^dk[[t]]\smallsetminus t^{d+1}k[[t]]$. Write $U=ct^d(1+w)$ with $w\in tk[[t]]$. Using the usual power series for $(1+X)^{1/n}$, we can write $(1+w)=(1+W)^n$ for some $W\in tk[[t]]$. Thus $U=x(t(1+W))^{d}$. There exists a continuous $k$-algebra automorphism of the local ring mapping $t$ to $t(1+W)$ (just because $t(1+W)$ has valuation 1). Hence after conjugation by this automorphism, we can assume that $w=0$, that is, $U=ct^d$.
Then we decompose $V$ according to the value modulo $d$ of the exponents: we write $V=\sum_{i=0}^{d-1}c_it^iP_i(ct^d)$$V=\sum_{i=0}^{d-1}t^iP_i(ct^d)$, with $c_i\in k$, $P_i(T)\in k[[T]]$. Thus $V$ belongs to a finite extension of $k((ct^d))$, and hence $V$ is algebraic over $k((ct^d))$. Hence there exists a nonzero $Q(X)\in k[[ct^d]][X]$ such that $Q(V)=0$. If we write $Q(X)=R(ct^d,X)$, we deduce $R(U,V)=0$, and $R$ is a nonzero element of $k[[X]][Y]\subset k[[X,Y]]$. (Of course the special form of $R$ is due to the fact we used an automorphism to assume $U$ is a monomial.)