= Solution
The <growth order of a meromorphic function> is
$$
\rho(f)=\limsup_{R\to\infty}\frac{\log^+T_f(R)}{\log R}.
$$
The order of the integrated count is defined analogously using $\log^+N(R;a)$, with an identically zero count assigned order zero. Constants have function order zero. These are growth orders, not local zero or <pole> orders. For nonconstant $f$, the first main theorem gives $\rho(N(\cdot;a))\le\rho(f)$ for every target.
Use the truncated <Nevanlinna second main theorem> in the following standard form: for distinct fixed targets $a_1,\ldots,a_q$,
$$
(q-2)T_f(R)\le\sum_{j=1}^q\overline N(R;a_j)+S_f(R),\qquad
S_f=O(\log^+T_f+\log R),
$$
outside an exceptional set of radii of finite linear measure. For finite order the error is $O(\log R)$; replacing each truncated count by its full count preserves the inequality.
Suppose $0<\rho(f)<\infty$ and three targets have counting orders below $\rho(f)$. Choose $\mu$ strictly between the maximum of their three orders and $\rho(f)$. Each full count is at most $R^\mu$ eventually. The theorem with $q=3$ then gives $T_f(R)\le C R^\mu$ for sufficiently large radii outside the exceptional set. Its tail measure is eventually less than one, so every interval $[R,R+1]$ contains a nonexceptional radius $R'$. Monotonicity of the characteristic yields $T_f(R)\le T_f(R')\le C(R+1)^\mu$ for all large $R$. This contradicts its order. If $\rho(f)=0$, all the counting orders are already zero. Hence the <counting-order exceptions for a finite-order meromorphic function> satisfy
$$
\boxed{\rho(N(\cdot;a))=\rho(f)\quad\text{apart from at most two targets}.}
$$
The exponential has order one but omits both zero and infinity, so those two counts are zero; this realizes two exceptions.
For a <Möbius map> $U(w)=(aw+b)/(cw+d)$ with $ad-bc\ne0$, translation and multiplication by a nonzero constant change $T$ by $O(1)$, as follows directly from the proximity inequalities and unchanged <pole> multiplicities. The first main theorem also gives $T(1/h)=T(h)+O(1)$. If $c=0$, $U$ is affine. If $c\ne0$, write $U(w)=a/c+K/(cw+d)$ with $K\ne0$ and use these three elementary operations. Thus $T_{U\circ f}=T_f+O(1)$, proving
$$
\boxed{\rho(U\circ f)=\rho(f).}
$$
For the sum, a <pole> of $f_1+f_2$ has order no larger than the sum of the two <pole> orders, and
$$
\log^+|f_1+f_2|\le\log2+\log^+|f_1|+\log^+|f_2|.
$$
Therefore $T_{f_1+f_2}\le T_{f_1}+T_{f_2}+O(1)$ and
$$
\boxed{\rho(f_1+f_2)\le\max(\rho(f_1),\rho(f_2)).}
$$
If $\rho(f_1)>\rho(f_2)$, apply the same inequality to $f_1=(f_1+f_2)-f_2$. It forces $\rho(f_1)\le\max(\rho(f_1+f_2),\rho(f_2))$, hence the sum's order equals $\rho(f_1)$. The reversed case is identical. This reasoning also handles an infinite larger order. For equal orders cancellation can be strict: $f_1=e^z$, $f_2=-e^z$ both have order one, but their sum is zero of order zero.
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