Why appending box dimensions to a sequence (in a loop) makes that sequence only contain duplicates of last dimension?
When you do:
\seq_put_right:Nn \mySeq { \dim_eval:n{ \box_wd:N \myBox } }
what you are appending to your sequence is the token list \dim_eval:n{\box_wd:N\myBox} (the tokens are \dim_eval:n, { with catcode 1, \box_wd:N, \myBox and } with catcode 2). The final \dim_eval:n is evaluated too late, once the loop has been fully executed and you do \seq_use:Nn \mySeq {,~}. At this point, \myBox contains the digit 9, hence what you saw. As egreg said (faster than me), replace \seq_put_right:Nn with \seq_put_right:Nx to evaluate \dim_eval:n inside the loop, when \myBox still contains the relevant digit.
You have to realize that your sequence variable \mySeq does not contain any length at all: it contains n times the same token list, which I explicitated above, with n=0 at first, then n=1, ..., and finally n=9. In other words, it contains n times the same “formula,” stored in the form of a token list. This formula is comparable to a function of one variable, the box variable \myBox. Whenever you use \seq_use:Nn \mySeq {...}, you put n copies of the same formula in the input stream (where n is the current length of \mySeq), and when TeX expands one of the \dim_eval:n tokens and processes all tokens resulting from this expansion, this yields a length (precisely: a 〈dimen〉) according to the formula and the contents of \myBox at that point.
At the end of your example, your code therefore computes n=9 times the same formula with the same value of \myBox, an \hbox containing the figure 9. In very explicit terms, your final \seq_use:Nn call is equivalent to this:
\dim_eval:n { \box_wd:N \myBox } ,~
\dim_eval:n { \box_wd:N \myBox } ,~
\dim_eval:n { \box_wd:N \myBox } ,~
\dim_eval:n { \box_wd:N \myBox } ,~
\dim_eval:n { \box_wd:N \myBox } ,~
\dim_eval:n { \box_wd:N \myBox } ,~
\dim_eval:n { \box_wd:N \myBox } ,~
\dim_eval:n { \box_wd:N \myBox } ,~
\dim_eval:n { \box_wd:N \myBox }
Calculating 9 times f(x) with the same function f and the same value of x gives 9 times the same result, hence the output you obtained.
For the \seq_use:Nn calls inside the loop, this is exactly the same, except that n is replaced by a number between 1 and 8, inclusive.
Note: as people already told you, your variables don't respect the LaTeX3 coding guidelines. Read expl3.pdf and l3styleguide.pdf from here (the beginning of interface3.pdf is also pretty useful to read at first, but I suppose you've already seen it).
You need to expand the value: use
\seq_put_right:Nx
instead.
What's the problem? If you do
\hbox_set:Nn \l_tmpa_box { A }
\seq_put_right:Nn \l_tmpa_seq { \dim_eval:n { \box_wd:N \l_tmpa_box } }
\hbox_set:Nn \l_tmpa_box { AA }
\seq_put_right:Nn \l_tmpa_seq { \dim_eval:n { \box_wd:N \l_tmpa_box } }
this results in the sequence to contain the following items (without the outer braces, as it would be displayed by \seq_show:N \l_tmpa_seq)
{ \dim_eval:n { \box_wd:N \l_tmpa_box } }
{ \dim_eval:n { \box_wd:N \l_tmpa_box } }
When the sequence is used, the width of the current contents of \l_tmpa_box is employed: the value it had at the moment of assignment has never been mentioned.
With \seq_put_right:Nx you instead get the value frozen at the moment of the assignment.
Would it be “better” to use
\exp_args:NNx \seq_put_right:Nn \l_tmpa_tl {...}
No. Using (or defining, in case of need) a variant is better style.