Computing a "two-sided p-value" when the binomial dist. is NOT symmetric When $p = .6$, then the binomial distribution is not symmetric. In this situation, how a "two-sided p-value" when observing 2 successes in 15 trials is calculated?
In R, such a p-value is obtained as follows:
binom.test(2, 15, p = .6, "two.sided")$p.value ## How this p-value has been calculated

Therefore, I'm asking what is the statistical basis for the calculation of 2-sided p-value in the above case?
A picture for my question:

 A: Exact p-value
The meaning of the p-value is the probability of getting the sample or more extreme than the sample under the null hypothesis. In your question, the distribution under the null hypothesis is Binomial with probability 0.6 on 15 trials. As you did on that graph, you already got the probability of 0, 1, 2,..., 15 successes among 15 trials. In your sample, you observed 2 successes. Next, comparing $\Pr(Y=x), x =0,1,2,...,15$ with $\Pr(Y=2)$. If $\Pr(Y=x) <= \Pr(Y=2)$, we consider $Y=x$ is extreme situation, so add $\Pr(Y=x)$ into your p-value. After going through all of the x, you get your exact p-value for the 2-side test.
A: Here is how I interpret your question: suppose a trial indicates a success probability of 60%, and we record 2 successes out of 15. We determine if this claim is valid.
Let $X$ be a random variable denoting these successes. Then $\bar{x} = 9$ and $\sigma = \frac{6}{\sqrt{10}}$. If we naively approximate a normal random variable for this data, we obtain a z-score of approximately $-3.689$ to which we obtain a p-value of $0.0002249$. We can reject our null hypothesis that the success probability is 60%.
If this is not the intent of the question, please edit the original post to include as much information as possible.
A: You can get a first approximation to the Clopper-Pearson interval by assuming the distribution is symmetric, in which case the events Y <= 2  are "just as extreme" as the events Y >= 15-2 = 13, and your approximate p-value is
p-value_1 = Pr(Y <= 2) + Pr( Y >= 13)
A glance at your picture reveals that symmetry wasn't a good assumption, as the event Y = 13 is quite a bit more likely than the event Y = 2.  Indeed the event Y=14 is also more likely than the event Y=2.   But the event Y=15 seems (from your picture) to be less likely than the event Y=2, so you might include it in your second-approximation to the p-value calculation (which should sum the probability of your observed event and of all "more extreme" events):
p-value_2 = Pr(Y <= 2) + Pr( Y >= 15)
For your third (and final) approximation, you'd check the probability density at Y=15.  I'm confident a_statistician got it right, with Pr(Y=15) < Pr(Y=2), so your two-tailed p-value is 
p-value = Pr(Y <= 2) + Pr( Y >= 16) 
         = Pr(Y <= 2)
As pointed out in Two-sided binomial test in Excel, the Clopper-Pearson 2-sided binomial test isn't something you'd want to perform in Excel.  You can hack your way through it for particular cases such as the one in your diagram.  Someday, someone somewhere will go to the trouble of developing and carefully-validating and publishing a Visual Basic routine BINOM.TEST.TWOSIDED.   But... why not learn S, or pay the licensing fees for SPSS?
A: This should get close.
Edit* This code below is Visual Basic for Applications which you would run in Developer mode in Excel.  It implements the solution outlined by a-statistician above.   First it calculates the probability (oP) of Hits out of trials with P probability.  Then, to find the extremes at each tail it cycles through all possible hits in trials and when it finds a probability (nP) that is more extreme (i.e., less than) the oP value in question, it adds it to the probability.
Function binom_test_tt(hits As Integer, trials As Integer, P As Double)

Dim i As Integer
Dim oP As Double
Dim nP As Double
Dim sum As Double

oP = Application.WorksheetFunction.Binom_Dist(hits, trials, P, False)

sum = oP

i = trials


While i <> -1  'cycle through all possible values

' get a new probability
nP = Application.WorksheetFunction.Binom_Dist(i, trials, P, False) 

'if more extreme than op, add it into the sum
 If (nP < oP) Then sum = sum + _ 
            Application.WorksheetFunction.Binom_Dist(i, trials, P, False) 

i = i - 1

Wend


binom_test_tt = sum


End Function

A: @a_statistician correctly points out to the fact that p-value refers to the probability of observed or extreme outcomes, given our null hypothesis is true.
Look, here our null hypotheses is  [Ho : p = 0.6]
And a two sided hypothesis test implies that our alternative hypothesis is 
[Ha: p != 0.1]
So as per the definition of p-value, our answer is simply the mathematical equation: we have our observed outcome to be p = 2.
p-value = P[k <= 2] + P[k >= 2], where k is my no. of successes. Apply the usual formula and you'll get at your answer.
