What can be shown is that the above expressions are equal in probability.
First check the distribution. As any linear combination of a Gaussian is Gaussian the right hand side is Gaussian - the left hand side too. Then we need the 2 moments:
The expected values - it is zero ... easy to see.
Next what you did not specify is that the correlation between $dW_t^1$ and $dW_t^2$ is $\rho$ then the variance can be calculated by
$$
VAR[\sigma_1dW_t^1+\sigma_2dW_t^2] = \sigma_1^2 VAR[dW_t^1] + 2 \sigma_1 \sigma_2 Covar[dW_t^1,dW_t^2] + \sigma_2^2 VAR[dW_t^2]
$$
which equals
$$
\sigma_1^2 dt + 2 \sigma_1 \sigma_2 \rho dt + \sigma_2^2 dt.
$$
On the other hand the variance of the lhs:
$$
VAR[\sqrt{\sigma_1^2 + 2 \sigma_1 \sigma_2 \rho+ \sigma_2^2} dW_t] = (\sigma_1^2 + 2 \sigma_1 \sigma_2 \rho+ \sigma_2^2) VAR[dW_t]
$$
and this is
$$
(\sigma_1^2 + 2 \sigma_1 \sigma_2 \rho + \sigma_2^2) dt,
$$
exactly what we needed.