Yes, the coordinate vectors of $\mathbf{v}_1,\ldots,\mathbf{v}_n$ with respect to the matrix $C$ will be a basis for $\mathbb{R}^n$.
Note that you are abusing notation somewhat: if $B$ and $C$ are bases, then they aren't matrices; what you really want is to have a basis $\beta$, and $B$ is the matrix whose columns are the vectors in $\beta$; and another basis $\gamma$, and $C$ is the matrix whose columns are the vectors in $\gamma$.
From what you have: since $CS = B$, and the kernel of $B$ is zero, then so is the kernel of $S$: if $\mathbf{x}$ lies in the kernel of $S$, then $\mathbf{0} = C\mathbf{0} = C(S\mathbf{x}) = (CS)\mathbf{x} = B\mathbf{x}.$ But since the kernel of $B$ is zero, that means that $\mathbf{x}=\mathbf{0}$. So the kernel of $S$ is zero, which means that its columns are linearly independent, hence a basis. But the columns of $S$ are the coordinate vectors of the elements of $\beta$ written in terms of $\gamma$, which gives your result.
For your final question, yes, you are correct. Here's a general statement you can prove (using the same technique as above):
If $A$ is $n\times p$, $B$ is $p\times m$, and $C$ is $n\times m$ with $AB=C$, then $\mathrm{ker}(B)\subseteq\mathrm{ker}(C)$. That is, if $\mathbf{x}\in\mathbb{R}^m$ is in the kernel of $B$, then it is also in the kernel of $C$.
Another way of attacking the problem is to prove that $[\mathbf{v}_1]_{\gamma},\ldots,[\mathbf{v}_n]_{\gamma}$ are linearly independent directly. But this is easy: suppose that $\alpha_1[\mathbf{v}_1]_{\gamma} + \cdots + \alpha_n[\mathbf{v}_n]_{\gamma} = \mathbf{0}.$ Then we have: $\mathbf{0} = \alpha_1[\mathbf{v}_1]_{\gamma} + \cdots + \alpha_n[\mathbf{v}_n]_{\gamma} = [\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n]_{\gamma}.$ But this means that $\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n=\mathbf{0}$, and since $\mathbf{v}_1,\ldots,\mathbf{v}_n$ are linearly independent, that means $\alpha_1=\cdots=\alpha_n=0$.
In essence, it's the same argument as you have above, but playing directly with the vectors instead of the associated change-of-basis matrices.