Proving equivalence of two substitutions by induction
First, let us show that the two assumptions are necessary. Here is an example showing what goes wrong when $x = y$. Take $t = x$, $u = 1$, $v = 2$. We have
$$
x\{x := 1\}\{x := 2\} = 1\{x := 2\} = 1,
$$
whereas
$$
x\{x := 2\}\{x := 1\{x := 2\}\} =
2\{x := 1\} = 2.
$$
Next, here is an example showing what goes wrong when $x$ is not free in $v$. Take $t = y$, $x = 1$, $y = x$. We have
$$
y \{x := 1\}\{y := x\} = y\{y := x\} = x,
$$
whereas
$$
y \{y := x\}\{x := 1\{y := x\}\} =
x\{x := 1\} = 1.
$$
Now let's do the base case more carefully. The term $t$ is just a variable. We have to consider three possibilities: $t=x$, $t=y$, $t \neq x,y$.
If $t=x$ then $t\{x := u\} = u$, and so the left-hand side is $u\{y := v\} = u\{y := v\}$. On the other hand, $t\{y := v\} = x$ (since $x \neq y$), and so the right-hand side is $x\{x := u\{y := v\}\} = u \{y := v\}$.
If $t=y$ then $t\{x := u\} = y$ (since $y \neq x$), and so the left-hand side is $y\{y := v\} = v$. On the other hand, $t\{y := v\} = v$, and so the right-hand side is $v\{ x:= u\{y := v\} \} = v$, since $x$ doesn't appear freely in $v$.
Finally, if $t \neq x,y$ then both sides are just equal to $t$, since the substitutions have no effect.
Next, the rest of the proof. The proof for function application is essentially as you wrote it. Let us denote the left-hand side by $L(t)$ and the right-hand side by $R(t)$. If $t = t_1t_2$ then
$$
L(t) = L(t_1) L(t_2) \stackrel{(\ast)}= R(t_1) R(t_2) = R(t),
$$
where $(\ast)$ is due to the induction hypothesis.
Finally, we need to handle abstraction. Let $t = \lambda m.s$. Applying $\alpha$-conversion, we can assume without loss of generality that $m \neq x,y$. This implies that
$$
L(t) = \lambda m.L(s) = \lambda m.R(s) = R(t).
$$
To see why $\alpha$-conversion is necessary, try doing the proof when $m=x$ or $m=y$. The case $m=x$ works out fine, but you will encounter a problem when $m=y$.