Tensors are said to be invariant under change of basis transformation and only dependent on the vectors $x,y,z,\dots,w$ which form the $n$-linear form $f(x,y,z,\dots,w)$. However, if in a certain basis, we have the vector $(1,1,3,\dots,2)$ and we express it in another basis so the vector is, say, $(2,3,4,\dots,1)$ and then we calculate the $n$-linear form of these two vectors (assume for example, that $f(x,y,z,\dots,w)= x-y$), the result will not coincide (for the first vector, it will equal $1-1=0$ and for the second one, $2-3=-1$).
Does not that contradict the fact that tensors are invariant?
Okay. I will take the most basic bilinear form: the dot product $$g(x,y) = x\cdot y$$
It eats two vectors and gives you back one scalar.
For concreteness, let's say our vector space is $\mathbb{R}^{2}$.Now let's take two different bases. Say
$$A = \left\{\begin{bmatrix}2\\3\end{bmatrix},\begin{bmatrix}1\\2\end{bmatrix}\right\} = \{A_1,A_2\}$$
$$B = \left\{\begin{bmatrix}1\\3\end{bmatrix},\begin{bmatrix}0\\2\end{bmatrix}\right\} = \{B_1,B_2\}$$
Now, let's calculate the components of our bilinear form $g$ in the basis $A$. For this, we operate on every pair of elemenst of the basis. So, for example: $$g_{11} = A_1\cdot A_1 = 2^2+3^2 = 13$$ and so on. You can check that $g_{12} = g_{21} = 8$ and $g_{22} = 5$. We will denote the components with respect to the basis $A$ by latin indices, so what we have calculated so far are the components $g_{ij}$.
We do this again for the basis $B$. You can check that the components of $g$ are $g_{11} = 10$, $g_{12} = g_{21} = 6$ and $g_{22} = 4$. We will denote these components by greek indices, so we just calculated $g_{\mu\nu}$
Now take two arbitrary vectors $x$ and $y$. I will take $x=\begin{bmatrix}7\\8\end{bmatrix}$ and $y=\begin{bmatrix}9\\5\end{bmatrix}$ Again, we take the components of $x$ and $y$ with respect to $A$ (and denote them by latin indices) and the components of $x$ and $y$ with respect to $B$ (and denote them by greek indices).
You can check that
With respect to $A$, we have $x_1 = 6$, $x_2=-5$ and $y_1=13$, $y_2=-17$.
With respect to $B$, we have $x_1 = 7$, $x_2=-\textstyle\frac{13}{2}$ and $y_1=9$, $y_2=-11$.
NOW FINALLY
Let's perform our calculation in base $A$. We perform our index calculation using Einstein summation convention.
\begin{align*} g(x,y) &= g_{ij}x^{i}y^{j}\\ &\begin{array}{ccc} =&g_{1j}x^{1}y^{j} &+ g_{2j}x^{2}y^{j}\\ =&g_{11}x^{1}y^{1} &+ g_{21}x^{2}y^{1}\\ &+g_{12}x^{1}y^{2} &+ g_{22}x^{2}y^{2}\\ =&13\cdot 6\cdot 13 &+ 8\cdot (-5)\cdot 13\\ &+8\cdot 6\cdot (-17) &+ 5\cdot (-5)\cdot (-17)\\ =&103 \end{array} \end{align*}
We repeat the same calculation, but now using the components with respect to the basis $B$.
\begin{align*} g(x,y) &= g_{\mu\nu}x^{\mu}y^{\nu}\\ &\begin{array}{ccc} =&g_{1\nu}x^{1}y^{\nu} &+ g_{2\nu}x^{2}y^{\nu}\\ =&g_{11}x^{1}y^{1} &+ g_{21}x^{2}y^{1}\\ &+g_{12}x^{1}y^{2} &+ g_{22}x^{2}y^{2}\\ =&10\cdot 7\cdot 9 &+ 6\cdot (-\textstyle\frac{13}{2})\cdot 9\\ &+6\cdot 7\cdot (-11) &+ 4\cdot (-\textstyle\frac{13}{2})\cdot (-11)\\ =&103 \end{array} \end{align*}
So yes, the result is invariant.