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If we have a set of linearly independent vectors, can they be mapped to a set of linearly dependent vectors? and vice versa. for example,

Given T : V → W is a linear map where $T(v_1) = w_1,··· ,T(v_n) = w_n$ for some vectors $w_1,··· ,w_n ∈ W$.

I think if the vectors $w_1,...,w_n$ are linearly independent then it doesn't necessarily mean that the vectors $v_1,...,v_n$ must also be linearly independent.
If we let $v_1= c_1v_2+...+c_n-1v_n$ , for some scalars $c$, and similarly write the other vectors in $V$ as linear combinations of each other, then $v_1,...,v_n$ would be dependent but would still be sent to linearly independent vectors in $W$.

Am I on the right track or is there a flaw in my logic/math. Thanks!

To adress your first question: Yes, linearly independent vectors may be mapped to linearly dependent vectors, consider for example the constant $0$ map.
Nevertheless, if you require that the map is injective, linearly independent vectors are mapped to linearly independent vectors which is easy to check.

To adress the second question: Whenever you have linearly dependent vectors their images will be linearly dependent, too.

Proof: Let $v_1,dots,v_n$ be linearly dependent, i.e. we can find $lambda_1,dots,lambda_n$ with not all $lambda_i=0$ such that $lambda_1 v_1+dots + lambda_n v_n=0$.
Then for the images we have
$$
lambda_1 T(v_1)+dots + lambda_n T(v_n)=T(lambda_1 v_1+dots + lambda_n v_n)=T(0)=0.
$$
Hence, $T(v_1),dots,T(v_n)$ are linearly dependent.

Yes, we can map a set of linearly independent vectors to a set of linearly dependent vectors. Take for instance the linear map represented by the matrix
$$left[matrix1 & 0\ 0 & 1\ 1& 1right]$$
The (linearly independent) standard basis of $mathbbR^3$, consisting of $[1,0,0]$, $[0,1,0]$, $[0,0,1]$, gets mapped to the vectors $[1,0]$, $[0,1]$ and $[1,1]$, which obviously form a linearly dependent set of vectors in $mathbbR^2$.

Suppose that $c_1w_1+...+c_nw_n=0$.

ie suppose that $T(c_1v_1+...+c_1v_n)=0$ (by linearity and $T(v_i)=w_i$.

Now if T is injective, you have that $c_1v_1+...+c_1v_n=0$, which in turn implies that $c_i=0$ for all i.

so $w_1,...w_n$ is linearly independent too.