They are both correct in their own context. They are describing two different ways of model selection in different situations.
In general, when you are doing model selection and testing, your data is divided into three parts, training set, validation set and testing set. You use your training set to train different models, estimate the performance on your validation set, then select the model with optimal performance and test it on your testing set.
On the other hand, if you are using K-fold cross-validation to estimate the performance of a model, your data is then divided into K folds, you loop through the K folds and each time use one fold as testing(or validation) set and use the rest (K-1) folds as training set. Then you average across all folds to get the estimated testing performance of your model. This is what the Wikipedia page is referring to.
But keep in mind that this is for testing a specific model, if you have multiple candidate models and want to do model-selection as well, you have to select a model only with your training set to avoid this subtle circular logic fallacy. So you further divide your (K-1) folds 'training data' into two parts, one for training and one for validation. This means you do an extra 'cross-validation' first to select the optimal model within the (K-1) folds, and then you test this optimal model on your testing fold. In other words, you are doing a two-level cross-validation, one is the K-fold cross-validation in general, and within each cross-validation loop, there is an extra (K-1)-fold cross-validation for model selection. Then you have what you stated in your question, 'Of the k subsamples one subsample is retained as the validation data, one other subsample is retained as the test data, and k-2 subsamples are used as training data.'
Best Answer
Usually the $k$-fold cross validation subsets have approximately equal size. It is just crucial that they don't overlap.
For example I just had a look at what WEKA does. Say that you have $N$ instances and $k$ folds, then $$ r = N \mod k $$ (the remainder of $N$ divided by $k$) is the number of surplus records. The first $r$ partitions will have $\lfloor N/k \rfloor + 1$ records, the other ones just $\lfloor N/k \rfloor$ instead
Regarding your example: $$N = 76 $$ $$k = 10 $$ $$ r = N \mod k = 6 $$
First $6$ partitions will have $ \lfloor N/k \rfloor + 1 = 7 + 1 = 8$ records, the other ones $ 7 $ instead.