What defines a tensor operator is the commutation relation of its components with the angular momenta. Thus, if $[L_a,K_b]=i\hbar\epsilon_{abc}K_c$ you have a vector operator by definition.

More generally you want something like \begin{align} [L_\pm,T^{\lambda}_\mu]&=\sqrt{(\lambda\mp\mu)(\lambda\pm \mu+1)}T^{\lambda}_{\mu\pm 1}\\ [L_0,T^{\lambda}_\mu]&=\mu T^{\lambda}_\mu\, , \end{align} which tends to be easier to work with and goes beyond vectors ($\lambda$ is not restricted to $1$).

It is a good bet that, if you start with a classical vector with components $x^a p_y^b$ with $a$ or $b=0$ or $1$ (or such types of products), the corresponding observable will also be a vector operator. If you have higher powers in $x$ or $p$ (i.e. both $a$ and $b\ge 2$), there may be ordering issues and you have to be quite careful. (You want to be aware that a product of hermitian operators remains hermitian only if the operators commute.)

Polynomials in $x,y,z$ alone, or in $p_x,p_y,p_z$ alone, can be decomposed into tensors by expressing them in terms of spherical harmonics. Thus, \begin{align} x+iy&= r \sin\theta e^{i\phi} = c r Y^1_{-1}(\theta,\phi)\, ,\\ (x+iy)(x-iy)&= a r^2 + b r^2 Y^2_{0}(\theta,\phi)\, ,\\ \end{align} for some constants $a,b,c$ and are thus linear combinations of tensor operators, with the latter containing a scalar and the component of a tensor with $\lambda=2$.

What defines a tensor operator is the commutation relation of its components with the angular momenta. Thus, if $[L_a,K_b]=i\hbar\epsilon_{abc}K_c$ you have a vector operator by definition.

More generally you want something like \begin{align} [L_\pm,T^{\lambda}_\mu]&=\sqrt{(\lambda\mp\mu)(\lambda\pm \mu+1)}T^{\lambda}_{\mu\pm 1}\\ [L_0,T^{\lambda}_\mu]&=\mu T^{\lambda}_\mu\, , \end{align} which tends to be easier to work with and goes beyond vectors ($\lambda$ is not restricted to $1$). It is also useful to use spherical components where, for instance, $$ T^1_1=-\frac{V_x+iV_y}{\sqrt{2}} $$

It is a good bet that, if you start with a classical vector with components $x^a p_y^b$ with $a$ or $b=0$ or $1$ (or such types of products), the corresponding observable will also be a vector operator. If you have higher powers in $x$ or $p$ (i.e. both $a$ and $b\ge 2$), there may be ordering issues and you have to be quite careful. (You want to be aware that a product of hermitian operators remains hermitian only if the operators commute.)

Polynomials in $x,y,z$ alone, or in $p_x,p_y,p_z$ alone, can be decomposed into tensors by expressing them in terms of spherical harmonics. Thus, \begin{align} x+iy&= r \sin\theta e^{i\phi} = c r Y^1_{-1}(\theta,\phi)\, ,\\ (x+iy)(x-iy)&= a r^2 + b r^2 Y^2_{0}(\theta,\phi)\, ,\\ \end{align} for some constants $a,b,c$ and are thus linear combinations of tensor operators, with the latter containing a scalar and the component of a tensor with $\lambda=2$.

In the case of more sophisticated products like the ones you have, you need to use the definition of composite tensors to properly combine the individual vectors using Clebsch-Gordan technology. Thus, again using spherical coordinates: $$ Z^{j}_m =\sum_{qq'} T^{\lambda}_q W^{\lambda'}_{q'} C_{\lambda,q;\lambda'q'}^{jm} $$ so it then becomes a matter of checking if the complicated expressions are properly combined using CGs (again note the use of spherical components).

A good source for this is the text by

Baym, Gordon. Lectures on quantum mechanics. CRC Press, 2018.