The time of one half-cycle of the classic two-transistor astable multivibrator is the time that it takes the base end of the capacitor to charge from (VCC (VBE + VCE(SAT))) to +VBE.
For example, at the moment right before the left-hand transistor switches on, the capacitor connected to its collector is charged to (nearly) VCC, with its left end positive. Note that the other end of the capacitor is held one diode drop above ground by the B-E junction of the other transistor. Now, immediately after the left-hand transistor switches on, it is now clamping the left end of that same capacitor to ground (actually, Vce(sat) above ground). Since the voltage across the capacitor can't change instantaneously, that means that the right end of it is initially driven to VCC. This is the starting point of the exponential curve for this half of the timing cycle.
Now, keep in mind that the capacitor is charging "toward" +VCC, but it gets halted by the B-E junction of the transistor at +VBE. This charging is occuring at a rate determined by the time constant C × R, and we're basically interested in the time that it takes to move halfway from its starting value to its final value. This works out to ln(0.5), or 0.693 times the R-C time constant.
For a more complete explanation of this circuit, see the Circuit Cellar "Engineering Quotient" column for issue 262 (April 2012).