A tensor on a manifold ($\mathbb{R}^3$ in your case) is a coordinate independent object. It is defined as a multilinear map $$T:\mathcal{T}_p\mathcal{M}^n\to\mathbb{R}$$ where $\mathcal{T}_p\mathcal{M}$ is the tangent space of the manifold $\mathcal{M}$ at a point $p$. I would like to answer your question with a short introduction on the concept of coordinate independence in differential geometry.

A manifold can locally be equipped with coordinates $\{x^\mu\}$ such that it locally always looks like a subset of $\mathbb{R}^n$ (if $\mathcal{M}\equiv\mathbb{R}^n$, there are still many different choices of coordinate systems). One can show that the space of directional derivatives fulfills the requirements of a vector space, this space is called the tangent space $\mathcal{T}_p\mathcal{M}$. A vector in this space is a directional derivative that acts on functions $f\in\mathcal{C}^1(\mathcal{M})$, meaning $$D_v(f)=v^\mu\partial_\mu f$$ Instead of $D_v$ one usually just writes $v$. The coordinate system $\{x^\mu\}$ suggests a basis of $\mathcal{T}_p\mathcal{M}$, namely $\{\partial_\mu\equiv\frac{\partial}{\partial x^\mu}\}$. Therefore $v=v^\mu\partial_\mu$. The $v^\mu$ are just a set of numbers which the physicist usually calls "vector", though they are really only the components of a vector. They key difference is that $v$ is a coordinate independent object; it is a directional derivative and a direction in space is something physical! Its components are coordinate dependent, coordinate systems have no physical reality!

Consider a different coordinate system, $\{x^{\mu^\prime}\}$. The tangent space basis $\{\partial_\mu\}$ as we know fulfills $$\partial_\mu=\frac{\partial}{\partial x^\mu}=\frac{\partial x^{\mu^\prime}}{\partial x^\mu}\frac{\partial}{\partial x^{\mu^\prime}}=\frac{\partial x^{\mu^\prime}}{\partial x^\mu}\partial_{\mu^\prime}$$ We call $J^{\mu^\prime}_{\phantom{\mu}\mu}=\frac{\partial x^{\mu^\prime}}{\partial x^\mu}$ the Jacobian matrix of the coordinate change. If $v$ is supposed to be a coordinate independent object, we can now find out how its components must transform: $$v^{\mu^\prime}\partial_{\mu^\prime}=v=v^\mu\partial_\mu=v^\mu J^{\mu^\prime}_{\phantom{\mu}\mu}\partial_{\mu^\prime}\implies v^{\mu^\prime}=J^{\mu^\prime}_{\phantom{\mu}\mu}v^\mu$$ This is what a physicist means by saying "a set of numbers $v^\mu$ transforms like a vector"! The physical object $v$ is unaffected by this!

We are missing one final step to get the final answer to your question: There exists a way of turning vectors into scalars. The dual space $\mathcal{T}^*_p\mathcal{M}$ is defined to be the set of linear maps $$\omega:\mathcal{T}_p\mathcal{M}\to\mathbb{R}$$ so if $v$ is a vector and $\omega$ is a dual vector, then $\omega(v)$ is a real number. It can again be expressed in coordinates, $$\omega(v)=\omega(v^\mu\partial_\mu)\overset{\text{linearity of }\omega}{=}v^\mu\omega(\partial_\mu)\equiv v^\mu\omega_\mu$$ Again, we call $\omega_\mu$ the (coordinate dependent) components of the dual vector (also called $1$-form) $\omega$. And again, even though we expressed it in coordinates, $\omega$ is a coordinate independent object, meaning that $\omega(v)$ should give the same real number no matter what coordinate system we used. In formula: $$v^{\mu}\omega_{\mu}=v^{\mu^\prime}\omega_{\mu^\prime}\overset{\text{transformation rule for }v^{\mu}}{=}v^\mu J^{\mu^\prime}_{\phantom{\mu}\mu}\omega_{\mu^\prime}\implies \omega_\mu=J^{\mu^\prime}_{\phantom{\mu}\mu}\omega_{\mu^\prime}$$ Notice how the transformation rule for $\omega$ is in some sense inverse to the transformation rule of $v$. We can also write $$\omega_{\mu^\prime}=(J^{-1})^\mu_{\phantom{\mu}\mu^\prime}\omega_\mu$$

Finally, tensors are objects that do not only take 1 vector (as $\omega$ did), but many vectors (say $n$), and turn them into a scalar. We would call them a multilinear map (whereas $\omega$ before was just a linear map, it had only 1 argument). Therefore they do not only have 1 index labelling the components (like the $\mu$ in $\omega_\mu$), but many indices (for example $T_{\mu_1\mu_2\dots\mu_n}$). However, by exactly the same arguments as before, in order to be coordinate independent objects on the manifold, the tensor components must transform like $$T_{\mu_1^\prime\dots\mu_n^\prime}=(J^{-1})^{\mu_1}_{\phantom{\mu}\mu_1^\prime}\dots(J^{-1})^{\mu_n}_{\phantom{\mu}\mu_n^\prime}T_{\mu_1\dots\mu_n}$$

A tensor on a manifold ($\mathbb{R}^3$ in your case) is a coordinate independent object. It is defined as a multilinear map $$T:\mathcal{T}_p\mathcal{M}^n\to\mathbb{R}$$ where $\mathcal{T}_p\mathcal{M}$ is the tangent space of the manifold $\mathcal{M}$ at a point $p$. I would like to answer your question with a short introduction on the concept of coordinate independence in differential geometry.

A manifold can locally be equipped with coordinates $\{x^\mu\}$ such that it locally always looks like a subset of $\mathbb{R}^n$ (if $\mathcal{M}\equiv\mathbb{R}^n$, there are still many different choices of coordinate systems). One can show that the space of directional derivatives fulfills the requirements of a vector space, this space is called the tangent space $\mathcal{T}_p\mathcal{M}$. A vector in this space is a directional derivative that acts on functions $f\in\mathcal{C}^1(\mathcal{M})$, meaning $$D_v(f)=v^\mu\partial_\mu f$$ Instead of $D_v$ one usually just writes $v$ ($D_v$ might actually denote a different type of derivative, the covariant derivative...that's another story). The coordinate system $\{x^\mu\}$ suggests a basis of $\mathcal{T}_p\mathcal{M}$, namely $\{\partial_\mu\equiv\frac{\partial}{\partial x^\mu}\}$. Therefore $v=v^\mu\partial_\mu$. The $v^\mu$ are just a set of numbers which the physicist usually calls "vector", though they are really only the components of a vector. They key difference is that $v$ is a coordinate independent object; it is a directional derivative and a direction in space is something physical! Its components are coordinate dependent, coordinate systems have no physical reality!

Consider a different coordinate system, $\{x^{\mu^\prime}\}$. The tangent space basis $\{\partial_\mu\}$ as we know fulfills $$\partial_\mu=\frac{\partial}{\partial x^\mu}=\frac{\partial x^{\mu^\prime}}{\partial x^\mu}\frac{\partial}{\partial x^{\mu^\prime}}=\frac{\partial x^{\mu^\prime}}{\partial x^\mu}\partial_{\mu^\prime}$$ We call $J^{\mu^\prime}_{\phantom{\mu}\mu}=\frac{\partial x^{\mu^\prime}}{\partial x^\mu}$ the Jacobian matrix of the coordinate change. If $v$ is supposed to be a coordinate independent object, we can now find out how its components must transform: $$v^{\mu^\prime}\partial_{\mu^\prime}=v=v^\mu\partial_\mu=v^\mu J^{\mu^\prime}_{\phantom{\mu}\mu}\partial_{\mu^\prime}\implies v^{\mu^\prime}=J^{\mu^\prime}_{\phantom{\mu}\mu}v^\mu$$ This is what a physicist means by saying "a set of numbers $v^\mu$ transforms like a vector"! The physical object $v$ is unaffected by this!

We are missing one final step to get the final answer to your question: There exists a way of turning vectors into scalars. The dual space $\mathcal{T}^*_p\mathcal{M}$ is defined to be the set of linear maps $$\omega:\mathcal{T}_p\mathcal{M}\to\mathbb{R}$$ so if $v$ is a vector and $\omega$ is a dual vector, then $\omega(v)$ is a real number. It can again be expressed in coordinates, $$\omega(v)=\omega(v^\mu\partial_\mu)\overset{\text{linearity of }\omega}{=}v^\mu\omega(\partial_\mu)\equiv v^\mu\omega_\mu$$ Again, we call $\omega_\mu$ the (coordinate dependent) components of the dual vector (also called $1$-form) $\omega$. And again, even though we expressed it in coordinates, $\omega$ is a coordinate independent object, meaning that $\omega(v)$ should give the same real number no matter what coordinate system we used. In formula: $$v^{\mu}\omega_{\mu}=v^{\mu^\prime}\omega_{\mu^\prime}\overset{\text{transformation rule for }v^{\mu}}{=}v^\mu J^{\mu^\prime}_{\phantom{\mu}\mu}\omega_{\mu^\prime}\implies \omega_\mu=J^{\mu^\prime}_{\phantom{\mu}\mu}\omega_{\mu^\prime}$$ Notice how the transformation rule for $\omega$ is in some sense inverse to the transformation rule of $v$. We can also write $$\omega_{\mu^\prime}=(J^{-1})^\mu_{\phantom{\mu}\mu^\prime}\omega_\mu$$

Finally, tensors are objects that do not only take 1 vector (as $\omega$ did), but many vectors (say $n$), and turn them into a scalar. We would call them a multilinear map (whereas $\omega$ before was just a linear map, it had only 1 argument). Therefore they do not only have 1 index labelling the components (like the $\mu$ in $\omega_\mu$), but many indices (for example $T_{\mu_1\mu_2\dots\mu_n}$). However, by exactly the same arguments as before, in order to be coordinate independent objects on the manifold, the tensor components must transform like $$T_{\mu_1^\prime\dots\mu_n^\prime}=(J^{-1})^{\mu_1}_{\phantom{\mu}\mu_1^\prime}\dots(J^{-1})^{\mu_n}_{\phantom{\mu}\mu_n^\prime}T_{\mu_1\dots\mu_n}$$