Set $m=n-k$.

In case $m=1$ the answer is YES and I hope you know it.

In general, you have $m\cdot k$ equations and $n=m+k$ unknowns. I.e. if $m\ge 2$ and $k>2$ or if $m> 2$ and $k\ge 2$ then the system is overdetermined. I think it should be possible to show that there are no solutions in this case.

I do not see what happes in the case $m=k=2$; i.e., for 2-dimensional submanifold in $\mathbb R^4$; maybe there is a solution for any $M$. In this case, if $\phi$ is a solution for generic $M$ then for fixed $x$ the points $\phi(x,y)$ do not lie in the normal plane to $\phi(x,0)\in M$; i.e. tubular-neighborhood-construction is useless here.

Set $m=n-k$.

If $m=1$ or $k=1$ then the answer is YES and I hope you know it.

In general, you have $m\cdot k$ equations and $n=m+k$ unknowns. I.e. if $m\ge 2$ and $k>2$ or if $m> 2$ and $k\ge 2$ then the system is overdetermined. I think it should be possible to show that there are no solutions in this case.

I do not see what happes in the case $m=k=2$; i.e., for 2-dimensional submanifold in $\mathbb R^4$; maybe there is a solution for any $M$. In this case, if $\phi$ is a solution for generic $M$ then for fixed $x$ the points $\phi(x,y)$ do not lie in the normal plane to $\phi(x,0)\in M$; i.e. tubular-neighborhood-construction is useless here.

Set $m=n-k$. In case $m=1$ the answer is YES and I hope you know it.

In general, you have $m\cdot k$ equations and $n=m+k$ unknowns. I.e. if $m\ge 2$ and $k>2$ or if $m> 2$ and $k\ge 2$ then the system is overdetermined. I think it should be possible to show that there are no solutions in this case.

I do not see what happes in the case $m=k=2$; i.e., for 2-dimensional submanifold in $\mathbb R^4$; maybe there is a solution for any $M$. But the tubular neighborhood is not good here.

Set $m=n-k$.

In case $m=1$ the answer is YES and I hope you know it.

In general, you have $m\cdot k$ equations and $n=m+k$ unknowns. I.e. if $m\ge 2$ and $k>2$ or if $m> 2$ and $k\ge 2$ then the system is overdetermined. I think it should be possible to show that there are no solutions in this case.

I do not see what happes in the case $m=k=2$; i.e., for 2-dimensional submanifold in $\mathbb R^4$; maybe there is a solution for any $M$. In this case, if $\phi$ is a solution for generic $M$ then for fixed $x$ the points $\phi(x,y)$ do not lie in the normal plane to $\phi(x,0)\in M$; i.e. tubular-neighborhood-construction is useless here.