Your question raises small questions of terminology and more interesting questions of how to think about data. I will stick with your question and focus on discrete variables. Most of what I say carries over to continuous variables, but with some need for re-wording and/or some differences in procedures.
First off, the mode is usually introduced and defined as just the most common value, namely the value with the highest frequency. That is the strictest sense of the term mode. When data are discrete, we look at frequencies and identify the value with the highest frequency.
When there are ties for the highest frequency, then so also there are ties for mode. With these invented data
value frequency
0 1
1 42
2 1
3 1
4 42
5 1
6 1
there is clearly a tie, with two modes at 1 and 4. However, had the frequencies been 42 and 41 it's my guess that most experienced users of statistics would still say that there were two modes, regardless of the rule that the mode is the value with the highest frequency. So, it is also true that a mode is a value with a pronounced peak in a frequency distribution, i.e. with a frequency notably higher than neighbouring values. (It's possible and common for a mode to be either the minimum or the maximum.)
Don't ask for a precise rule, or even rule of thumb, on what counts as pronounced or notable; it's what is obvious when you graph it and the decision comes quickly with a little experience.
The importance and interest of modes often lies in what they indicate, which is sometimes that there are qualitatively different groups being mixed together in the sample, such as men and women or healthy and sick people. Sometimes there are physical reasons for having two modes. In some climates, there are two common states of cloudiness of the sky, almost cloud-free days and clouded-over days.
I've not seen the term modality being used except occasionally as meaning the number of modes. Statistical people certainly talk about bimodality, meaning that the data are bimodal, or have two modes; or multimodality, meaning that the data are multimodal, or have many modes. Some of these terms are a little unnecessary and arguably relict from a time when there was a stronger inclination among scholars and scientists to invent words based on Latin and Greek roots, but they are quite often used.
The second part of your question I read as asking whether the median should be computed as the average of two modes. I may be missing something here, but I guess you are mixing in a quite different question. The computation of medians has nothing to do with modes at all. It's just the convention that with an even number of values, you should report the median as the mean of the two middle values. That's a convention, but it is taught in introductory statistics courses as a rule to be followed. With grouped data, the principle is still the same. It's quite possible with discrete data that interpolation will cause the median to be reported as a value that is not observable, e.g. 2.5 children. That's not something to worry about.
Back to terminology: I'd assert that modality is not another word for mode. Still less can it be used to refer to any value.
EDIT: I tried to pitch my original answer in a way that should help others apart from the OP. I focused on what seemed the more interesting question of a mode is, and downplayed a question about the median which seems confused on a point well covered in just about every elementary text. I've not tried to keep pace with repeated edits of the original question with more emphasis on how to compute medians.
I have read a little bit that refers to a Wilcoxon-Mann-Whitney test for difference in medians
The Wilcoxon-Mann-Whitney is not a test for equality of medians, it's a test for one variable being stochastically larger than another ($P(X>Y)>P(Y>X)$). If you're using it as a location test, it's actually a test for a zero (population) median of pairwise differences.
If you make the additional assumption of identical distributions apart from a location shift then it will be a test of medians (but it would also be a test of means, as long as population means exist).
I have a survey where respondents were asked to rank a series of statements in terms of how much they agreed with them. There were seven statements. So, for each of seven statements, each respondent has value that ranges from 1 to 7, depending on where they ranked that item.
So that's seven numbers per respondent. What did you do with those 7 numbers?
first, should I conceive of this as an ordinal or an interval level variable.
If you just added the scores on each item, you already assumed the 7 items were interval scale when you did that.
because I'm not sure how meaningful a mean is in this context.
If you can add a "2" and "6" and get the same result as when you got a "3" and "5", then everything needed for a mean to be meaningful was already assumed to be true.
It seems more intuitive to say the median ranking of this group of respondents' for this statement was 2, compared to 5 for a different group, rather than, say, 3.6 compared to 5.5, or whatever.
If you find a median intuitive, that's fine, there's nothing stopping you working with medians -- that doesn't require you to assume it ordinal even though you already made it interval.
You can always consider a permutation test; it assumes exchangeability under the null, which is a somewhat stronger assumption than the Wilcoxon-Mann-Whitney needs.
Best Answer
Medians are not linear, so there are a variety of circumstances under which something like that (i.e. $\text{median}(X_1)+\text{median}(X_2)<\text{median}(X_1+X_2)$) might happen.
It's very easy to construct discrete examples where that sort of thing occurs, but it's also common in continuous situations.
For example it can happen with skewed continuous distributions - with a heavy right tail, the medians might both be small but the median of the sum is "pulled up" because there's a good chance that one of the two is large, and a value above the median is typically going to be far above it, making the median of the sum larger than the sum of the medians.
Here's an explicit example: Take $X_1,X_2 \, \stackrel{\text{i.i.d.}}{ \sim} \operatorname{Exp}(1)$. Then $X_1$ and $X_2$ have median $\log(2) \approx 0.693$ so the sum of the medians is less than $1.4$, but $X_1+X_2\sim \operatorname{Gamma}(2,1)$ which has median $\approx 1.678$ (actually $ -W_{-1}(-\frac{1}{2 e}) - 1$ according to Wolfram Alpha)