You are right, and here is why. $|0,1\rangle$ and $|1,0\rangle$ form a basis in the Hilbert space $H$, which has two complex dimensions. It can be the space of one boson which may be in two quantum states. The state space for two bosons will be not merely the tensor product of the space $H$ with itself, $H\otimes H$. It will be a subspace of it, which is denoted sometimes by $H\odot H$, and is the symmetric tensor product. That is, in general, while the tensor product of two vectors $u$ and $v$ is $u\otimes v$, the symmetric tensor product is $u\odot v=\frac{1}{\sqrt {2}}\left(u\otimes v + v \otimes u\right)$. The factor $\frac{1}{\sqrt {2}}$ is chosen for normalization. The complex dimension of $H$ is $2$, and that of $H\otimes H$ is $4$. But the complex dimension of $H\odot H$ is $3$. The basis of $H\otimes H$ is given by $|0,1\rangle|0,1\rangle$ ($ :=|0,1\rangle\otimes|0,1\rangle$, but it is customary to omit $\otimes$), $|0,1\rangle|1,0\rangle$, $|1,0\rangle|0,1\rangle$, $|1,0\rangle|1,0\rangle$. But the basis of $H\odot H$ is $|0,2\rangle:=|0,1\rangle|0,1\rangle$, $|1,1\rangle:=\frac{1}{\sqrt {2}}\left(|0,1\rangle|1,0\rangle+|1,0\rangle|0,1\rangle\right)$, and $|2,0\rangle:=|1,0\rangle|1,0\rangle$. So you are right that $|0,2\rangle=|0,1\rangle|0,1\rangle$, and that $|1,1\rangle=\frac{1}{\sqrt {2}}\left(|0,1\rangle|1,0\rangle+|1,0\rangle|0,1\rangle\right)$, and that the state $|1,1\rangle$ is independent on the chosen basis.
I add this to explain in more detail some points which were raised. The Fock space for a boson whose one-particle Hilbert space is $H$ is
$$F=\bigoplus_{k=0}^{\infty}\odot^kH=\mathbb C\oplus H \oplus \left(H\odot H\right)\oplus\ldots\oplus\odot^kH\oplus\ldots.$$
Now we see that $H$ is a subspace of $F$, or $H\lt F$. Tensor products between $H$ and $H$ don't stay in general in $F$, because $H\otimes H\nleq F$. But $F$ is closed to symmetric tensor products, that is $H\odot H<F$.
So, it is perfectly legit to take tensor products inside the Fock space, provided that they are symmetric tensor products.
The full basis of the Fock space, in the full notation involving tensor products and direct sums of vector spaces, is complicated to write. So, rewrite it like this:
$$|n_1,\ldots,n_m\rangle:=|1,0,\ldots,0\rangle^{n_1}\odot|0,1,\ldots,0\rangle^{n_2}\odot\ldots\odot|0,0,\ldots,1\rangle^{n_m}$$
where $m=\dim H$, and the exponent in $|1,0,\ldots,0\rangle^{n_1}$ means $n_1$ times tensor product with itself (it is automatically symmetric).
Now go back to $\dim H=m=2$. The basis will be
$$|n_1,n_2\rangle=|1,0\rangle^{n_1}\odot |0,1\rangle^{n_2}.$$
If you check this for $n_1+n_2=2$, you will find exactly what I said in the answer.