The advantage of cartesian coordinates is that we can take vectors as simply indexed functions. But there is an additional feature that can not be forgotten when changing variables: coordinate basis.

The velocity vector of the OP example can be expressed in its complete form as:

$v_1(X^1, X^2) = V^1B_{11} + V^2B_{21}$
$v_2(X^1, X^2) = V^1B_{12} + V^2B_{22}$

In cartesian coordinates, $X^1 = x$, $X^2 = y$, $V^1 = v_x$, $V^2 = v_y$, $B_{11} = 1$, $B_{12} = 0$, $B_{21} = 0$, $B_{22} = 1$

That way:
$v_1(x,y) = V^1$ and $v_2(x,y) = V^2$ are the familiar cartesian components of the indexed function (and vector) $v(x,y)$.

When transforming to polar coordinates, it is possible to derive $B_{ab}$ and $V^a$ such that $v_b$ don't change. Here:
  $B_{11} = cos(\theta)$, $B_{12} = sin(\theta)$, $B_{21} = -Rsin(\theta)$, $B_{22} = Rcos(\theta)$, $X^1 = R$, $X^2 = \theta$, $V^1 = v_R$, $V^2 = v_{\theta}$

So:
$v_1(R, \theta) = v_Rcos(\theta) - v_{\theta}Rsin(\theta)$
$v_2(R, \theta) = v_Rsin(\theta) + v_{\theta}Rcos(\theta)$

In the example, $v_1 = v_0$ and $v_2 = 0$. It is easy to realize that to keep the same values:

$$v_R = v_0cos(\theta)$$ and $$v_\theta = -v_0\frac{sin(\theta)}{R}$$

In polar coordinates, the components change with time for this vector be constant with time as desired. The vector equations of the Newton laws are valid, but the notion of what is a vector must be closely understood.

The advantage of cartesian coordinates is that we can take vectors as simply indexed functions. But there is an additional feature that can not be forgotten when changing variables: coordinate basis.

The velocity vector of the OP example can be expressed in its complete form as:

\begin{align} v_1(X^1, X^2) &= V^1B_{11} + V^2B_{21}\\ v_2(X^1, X^2) &= V^1B_{12} + V^2B_{22}. \end{align}

In cartesian coordinates, \begin{align} X^1 = x, &\quad X^2 = y \\ V^1 = v_x, &\quad V^2 = v_y \\ B_{11} = 1, &\quad B_{12} = 0 \\ B_{21} = 0, &\quad B_{22} = 1. \end{align}

That way:
$v_1(x,y) = V^1$ and $v_2(x,y) = V^2$ are the familiar cartesian components of the indexed function (and vector) $v(x,y)$.

When transforming to polar coordinates, it is possible to derive $B_{ab}$ and $V^a$ such that $v_b$ don't change. Here: \begin{align} B_{11} = \cos(\theta), &\quad B_{12} = \sin(\theta) \\ B_{21} = -R \sin(\theta), &\quad B_{22} = R \cos(\theta) \\ X^1 = R, &\quad X^2 = \theta \\ V^1 = v_R, &\quad V^2 = v_{\theta}. \end{align}

So:
\begin{align} v_1(R, \theta) &= v_R \cos(\theta) - v_{\theta}R \sin(\theta)\\ v_2(R, \theta) &= v_R \sin(\theta) + v_{\theta}R \cos(\theta). \end{align}

In the example, $v_1 = v_0$ and $v_2 = 0$. It is easy to realize that to keep the same values:

$$v_R = v_0 \cos(\theta) \quad \text{and} \quad v_\theta = -v_0 \frac{\sin(\theta)}{R}.$$

In polar coordinates, the components change with time for this vector be constant with time as desired. The vector equations of the Newton's laws are valid, but the notion of what is a vector must be closely understood.