1 ) When B travels towards A, wouldn't each twin observe the other's clock moving ["running", "ticking"] slower?
No, quite the opposite:
If two participants (such as A and B) were moving towards each other (at constant mutual speed, $v := c~\beta$, $\beta > 0$)
and if one stated signal indications at constant frequency $f$ (as "ticks" of a good clock)
then the indications of the other having observed these signal "ticks" occured at larger frequency $f \sqrt{\frac{1 + \beta}{1 - \beta}}$; or in other words: quicker than the signal indications had been stated; cmp. blueshift.
In the described setup this applies to both clocks, mutually.
let's say the twins start out some distance apart, in the same reference frame.
... more precisely: the twins are initially at rest to each other, some distance apart.
2) The initial positions of A and B contain synchronized clocks
Let's also suppose
- that both clocks had each been (initially) "ticking" at particular frequencies,
- that the synchronization established that these frequency of one had been equal to the frequency of the other, and
- that both clocks had kept (ticking at) their constant equal frequency throughout the experiment; i.e. the clocks were and remained good throughout the experiment.
3) At some pre agreed upon time B accelerates towards A, reaches some velocity and travels all the way to A at that velocity. [...]
when they meet and before A decelerates what would each one observe regarding the other's clock?
At their meeting A's clock indicates $n$ ticks after the synchronized "pre agreed upon start time", and B's clock indicates $n \sqrt{ 1 - \beta^2 }$ ticks after the synchronized "pre agreed upon start time";
or almost, approximately, as far as the "acceleration phase" B is of negligible duration compared to the "drift phase".
The important point is: B's duration from indicating the "pre agreed upon start time" (and starting to accelerate towards A) until indicating the meeting and passing of A is shorter by a factor of $\sqrt{ 1 - \beta^2 }$ than A's duration from indicating the "pre agreed upon start time" until indicating the meeting and passing of B.
The following sketch illustrates the case of $\beta = 0.6$, therefore with mutual blueshift factor $\sqrt{ \frac{1 + \beta}{1 - \beta} } = 2$:

When B decelerates and enters A's reference frame would he observe A spontaneously age?
When B decelerates and has come to rest with respect to A they can again determine which indication of one has been simultaneous to any indication of the other.
And both clocks would again tick at equal frequency, since both clocks were supposed to have been and to have remained good.
As far as the "deceleration phase" B, too, is of negligible duration compared to the "drift phase", they would still find the relation between their good clocks as it was observed at their meeting:
clock A is and remains $n (1 - \sqrt{ 1 - \beta^2 })$ ticks ahead of clock B.