[Carlo Lazzaro]

Ruby:
1) why you were advised to _test- the linear combination of B1+B3=0 I cannot say (unless is one of your research goals).
2) their linear combination is not statistical signigficant. That's all.

[Carlo Lazzaro]

Ruby:
Handling the issue from another side, -test- ing B1+B3=0, is like -test-ing B1=-B3.
Considering the following toy-example:

use "https://www.stata-press.com/data/r16/nlswork.dta"
.  areg ln_wage i.south##i.not_smsa, vce(cluster idcode) abs(idcode)

Linear regression, absorbing indicators         Number of obs     =     28,526
Absorbed variable: idcode                       No. of categories =      4,711
                                                F(   3,   4710)   =      12.06
                                                Prob > F          =     0.0000
                                                R-squared         =     0.6270
                                                Adj R-squared     =     0.5532
                                                Root MSE          =     0.3196

                               (Std. Err. adjusted for 4,711 clusters in idcode)
--------------------------------------------------------------------------------
               |               Robust
       ln_wage |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
---------------+----------------------------------------------------------------
       1.south |   -.070137   .0235179    -2.98   0.003    -.1162432   -.0240309
    1.not_smsa |  -.1188317   .0239774    -4.96   0.000    -.1658386   -.0718248
               |
south#not_smsa |
          1 1  |   .0691241   .0334817     2.06   0.039     .0034843    .1347639
               |
         _cons |    1.72652   .0102768   168.00   0.000     1.706373    1.746668
--------------------------------------------------------------------------------

. test 1.south + 1.south#1.not_smsa=0

 ( 1)  1.south + 1.south#1.not_smsa = 0

       F(  1,  4710) =    0.00
            Prob > F =    0.9753

. test 1.south =-(1.south#1.not_smsa)

 ( 1)  1.south + 1.south#1.not_smsa = 0

       F(  1,  4710) =    0.00
            Prob > F =    0.9753

.

As, regardless their signs, the magnitude of the two coefficients is similar, no wonder that you cannot reject the null that their linear combination =0.