Clock Tree Jitter Estimation
In general you cannot add RMS values, but you can add the squares of the values.  If you are familiar with standard deviation and variance in statistics, it is the same concept.

Calculating the resulting random RMS jitter (note, this is neglecting the deterministic contributor):

$$ \sqrt{(1.3 \text{ ps rms})^2 + (0.35 \text{ ps rms})^2} = 1.346 \text{ ps rms}$$

Peak-to-peak is somewhat more complicated. Statistically, if you run your test for long enough, the peak-to-peak jitter is unbounded. Are there any details in the oscillator datasheet about exactly how they define peak-to-peak jitter?  Or perhaps more importantly, what is the requirement that you are trying to meet?



Update 1 May 2017:

The comments here have inspired me to think about this a little more. Even without the datasheet for the clock oscillator, if we assume that they used some linear scaling factor to convert between pk-pk and rms jitter, then we should be able to figure out what that factor is.  The spec gives 

$$25.6\text{ ps p-p} = 1.3\text{ ps rms} + 6.7\text{ ps p-p}$$

Assuming that they scaled their rms value to get a p-p value, and using the formula to sum rms values:

$$\frac{25.6}{\alpha} = \sqrt{1.3^2 +\left(\frac{6.7}{\alpha}\right)^2} \implies \alpha = 19.006 $$

So they dividied their peak to peak jitter components by 19 to convert it to rms jitter when they combined their terms.  Thus, we can use the same value to convert their resulting J1 value back to rms, getting \$1.347\text{ ps rms}\$.

Now that we know the rms jitter of the oscillator, we can more accurately combine it with the buffer:

$$ \sqrt{(1.347 \text{ ps rms})^2 + (0.35 \text{ ps rms})^2} = 1.392 \text{ ps rms}$$

To convert that to pk-pk, you can use a calculator like this: https://www.jitterlabs.com/support/calculators/rms-peak-peak-calculator based on your signal frequency and acceptable error rate.  I plugged in some values and you are definitely below 40 ps pk-pk, assuming there are no other error sources such as from the power supply, as mentioned in another answer.