Rate and order confusion
Although Nicolau Saker Neto's answer provides the essential, I wish here to propose two emblematic examples for the situations cited in your question:

First case: two steps process, the first being the slowest.

Reaction between $\ce{H2O2}$ and $\ce{HBr}$:

\begin{align}
\ce{H2O2 + H+ +Br- &amp;-&gt; HOBr + H2O}  &amp;&amp; \text{slow}\\
\ce{HOBr + H+ +Br- &amp;-&gt; Br2 + H2O}   &amp;&amp; \text{fast}\\
\end{align}

The rate of this reaction, determined by the slow step, equals:

$$
\frac{d[\ce{Br2}]}{dt}=k_{1}\cdot[\ce{H2O2}]\cdot[\ce{H+}]\cdot[\ce{Br-}]
$$

This is therefore a third order reaction.  Stoichiometric coefficients of the slow elementary step yield the specific order in each of the three reacting species.

Second case: two steps process, the first being the fastest.

Reaction between $\ce{NO}$ and $\ce{O2}$:

The initial equilibrium is rapidly established:

\begin{align}
\ce{2NO &amp;&lt;=&gt; N2O2}      &amp;&amp; \text{fast}\\
\ce{N2O2 + O2 &amp;-&gt; 2NO2} &amp;&amp; \text{slow}\\
\end{align}

In this case, the rate is:

$$
\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot[\ce{N2O2}]\cdot[\ce{O2}]
$$

This is therefore a second order reaction, with specific orders coinciding, as usual, with the stoichiometric coefficients in the slow step. Using the equilibrium constant $K$ of the first step, one can finally express the rate of the global equation as a function of $[\ce{NO}]$ and $[\ce{O2}]$, say:

$$
\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot K \cdot[\ce{NO}]^{2}\cdot[\ce{O2}]
$$



Summarizing, as you correctly say, the stoichiometric coefficient of a given reactant, in the rate determining elementary step ("slow" step), directly yields the order by which the concentration of that reactant is raised in the rate equation.