It will turn out that the rotation in polar coordinates is nothing else than a shift
of coordinate functions by an angle $\alpha$. Note that we usually shift a function
$g$ on $\mathbb R$ by $\alpha$ by going to the function $x\mapsto g(x\color{red}{-}\alpha)$.
A very convenient method to formalize vector fields in different coordinates is to treat them as derivations. This is very common in differential geometry. In $2d$ Cartesian coordinates a vector field is then
$$
X=X_1\frac{\partial}{\partial x_1}+X_2\frac{\partial}{\partial x_2}
$$
where $X_1,X_2$ are scalar functions on $\mathbb R^2$.
From the Cartesian-polar transformation
\begin{align}
r&=\sqrt{x_1^2+x_2^2}\,,&
\theta&=\arctan\frac{x_2}{x_1}\,\\
x_1&=r\cos\theta\,,&x_2&=r\sin\theta
\end{align}
we get
\begin{align}
\frac{\partial x_1}{\partial r}&=\cos\theta\,,&\frac{\partial x_1}{\partial \theta}=-r\sin\theta\,,\\
\frac{\partial x_2}{\partial r}&=\sin\theta\,,&\frac{\partial x_2}{\partial \theta}=r\cos\theta\,.
\end{align}
By the chain rule we have for any differentiable function $f$ on $\mathbb R^2$
\begin{align}
\frac{\partial f}{\partial r}&=\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial r}+\frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial r}
=\frac{\partial f}{\partial x_1}\cos\theta+\frac{\partial f}{\partial x_2}\sin\theta\,,\\[3mm]
\frac{\partial f}{\partial \theta}&=\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial \theta}+\frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial \theta}
=-\frac{\partial f}{\partial x_1}r\sin\theta+r\frac{\partial f}{\partial x_2}\cos\theta\,.
\end{align}
In matrix notation and dropping the test function this is the transformation rule for the basis vector fields:
$$
\begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \theta} \end{pmatrix}=\begin{pmatrix}\cos\theta &\sin\theta\\-r\sin\theta&r\cos\theta\end{pmatrix}\begin{pmatrix}\frac{\partial}{\partial x_1}\\\frac{\partial}{\partial x_2} \end{pmatrix}.
$$
Clearly,
$$
\begin{pmatrix}\frac{\partial}{\partial x_1}\\\frac{\partial}{\partial x_2} \end{pmatrix}=\begin{pmatrix}\cos\theta &-\frac{1}{r}\sin\theta\\\sin\theta&\frac{1}{r}\cos\theta\end{pmatrix}\begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \theta} \end{pmatrix}.
$$
The vector field $X$ in polar coordinates is therefore,
\begin{align}
X'&=X_1'\Big(\cos\theta\frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial \theta}\Big)+X_2'\Big(\sin\theta\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\frac{\partial}{\partial \theta}\Big)\\[3mm]
&=\Big(X_1'\cos\theta+X_2'\sin\theta\Big)\frac{\partial}{\partial r}+
\Big(-X_1'\frac{1}{r}\sin\theta+X_2'\frac{1}{r}\cos\theta\Big)\frac{\partial}{\partial \theta}
\end{align}
where $X'_1(r,\theta)=X_1(x_2,x_2)$ and $X'_2(r,\theta)=X_2(x_2,x_2)\,.$
The whole point is now that nothing is simpler than rotating the vector field $X'$ by an angle $\alpha$ around the origin.
This rotation is a shift which gives
\begin{align}
X''
&=\Big(X_1''\cos(\theta-\alpha)+X_2'\sin(\theta-\alpha)\Big)\frac{\partial}{\partial r}+
\Big(-X_1''\frac{1}{r}\sin(\theta-\alpha)+X_2''\frac{1}{r}\cos(\theta-\alpha)\Big)\frac{\partial}{\partial \theta}
\end{align}
where $X''_1(r,\theta)=X'_1(r,\theta-\alpha)$ and $X''_2(r,\theta)=X_2'(r,\theta-\alpha)\,.$
