JavaScript (ES6), 106 104 100 bytes
Saved 2 bytes thanks to @Shaggy
a=>a.filter(n=>n,[...a].map((v,i)=>a[a[p>v&(n=~~a[i+1])<p?k:i+(k=i,n>v&p<n)]+=x=a[i],p=v,i]-=x,p=0))
Commented
We first update the original input array a[] by iterating on a copy of it. During this step, all values 'attracted' by other ones are set to \$0\$.
Because the array is parsed from left to right, we can just add \$a_i\$ to \$a_{i+1}\$ whenever a value is attracted by its right neighbor.
Example: \$4\rightarrow5\rightarrow6\$ is turned into \$[0,\color{red}9,6]\$ and then \$[0,0,\color{red}{15}]\$.
But when several values in a row are attracted by their left neighbor, we need to add \$a_i\$ to the first attractor \$a_k\$ of this sequence (with \$k<i\$) rather than simply \$a_{i-1}\$.
Example: \$6\leftarrow5\leftarrow4\$ is turned into \$[\color{red}{11},0,4]\$ and then \$[\color{red}{15},0,0]\$.
[...a] // create a copy of a[]
.map((v, i) => // for each value v in a[] at position i:
a[ // this statement updates a[i]:
a[ // this statement updates either a[i] or an adjacent value:
p > v & // if the previous value p is greater than v
(n = ~~a[i + 1]) // and the next value n
< p ? // is less than p (attraction to the left):
k // use k (k is initially undefined, but this code cannot
// be triggered during the first iteration)
: // else:
i + ( // use either i or i + 1:
k = i, // set k to i
n > v & // use i + 1 if n is greater than v
p < n // and p is less than n (attraction to the right)
) //
] += x = a[i], // add x = a[i] to the entry defined above
p = v, // update the previous value to v
i // actual index to update a[i]
] -= x, // subtract x from a[i]
p = 0 // start with p = 0
) // end of map()
We then filter out all entries equal to \$0\$.
a.filter(n => n)
